100 Integral Problems and Solutions (part 2)
Hai Guys..
Assalamu'alaikum Wr.Wb.
Nama saya Hasbiansyah Cahyadi
Disini saya akan melanjutkan pembahasan soal integral sebelumnya silahkan kalian untuk menyimak pembahasannya...
Berikut adalah 100 soal integral yang akan saya bahas
Pembahasan ini akan saya jadikan 4 bagian, untuk soal ini adalah problem no 26-50
Ini adalah pembahasan part ke 2, yaitu soal no 26-50
26) $\int\frac{x^2}{\sqrt{4-x^2}}\ dx$
Jawab : Misalkan $x=2sin\ y$
$dx=2cos\ y\ dy$
$\int\frac{x^2}{\sqrt{4-x^2}}\ dx$
=$\int\frac{4sin^2\ y}{2cos\ y}.2cos\ y\ dy$
=$\int 4sin^2\ y\ dy$
=$\int 2(1-cos\ 2y)\ dy$
=$\int 2\ dy-\int cos\ 2y\ dy$
=$2y-\frac{sin\ 2y}{2}+C$
=$2y-sin\ y.cos\ y+C$
=$2sin^{-1}\left(\frac{x}{2}\right)-\frac{x}{2}.\left(\frac{\sqrt{4-x^2}}{2}\right)+C$
=$2sin^{-1}\left(\frac{x}{2}\right)-\frac{x\sqrt{4-x^2}}{4}+C$
27) $\int\sqrt{2x-x^2}\ dx$
Jawab :
$\int \sqrt{2x-x^2}\ dx$
=$\int \sqrt{1-(1-x)^2}\ dx$
misalkan : $1-x=sin\ y$
$-dx=cos\ y\ dy$
=$\int\sqrt{1-sin^2\ y}.(-cos\ y)\ dy$
=$\int -cos^2\ y\ dy$
=$\int -\frac{cos\ 2y+1}{2}\ dy$
=$-\int\frac{cos\ 2y}{2}\ dy-\int\frac{1}{2}\ dy$
=$-\frac{sin\ 2y}{4}-\frac{y}{2}+C$
=$-\frac{sin\ y.cos\ y}{2}-\frac{y}{2}+C$
=$-\frac{(1-x)\sqrt{2x-x^2}}{2}-\frac{sin^{-1}(1-x)}{2}+C$
28) $\int\frac{4x-2}{x^3-x}\ dx$
Jawab :
$\int\frac{4x-2}{x(x-1)(x+1)}\ dx$
$\int\left(\frac{2}{x}+\frac{1}{x-1}+\frac{-3}{x+1}\right)\ dx$
=$2ln|x|+ln|x-1|-3ln|x+1|+C$
=$ln\left|\frac{x^2(x-1)}{(x+1)^3}\right|+C$
29) $\int\frac{x^4}{x^2-2}\ dx$
Jawab :
Perhatikan bahwa
$x^4=(x^2-2)^2+4x^2-4=(x^2-2)^2+4(x^2-2)+4$
$\int\frac{x^4}{x^2-2}\ dx$
=$\int (x^2-2)+4+\frac{4}{x^2-2}\ dx$
=$\int x^2+2\ dx+\int\frac{4}{x^2-2}\ dx$
Obs : $\int\frac{4}{x^2-2}\ dx$
=$\int\frac{4}{(x-\sqrt{2})(x+\sqrt{2})}\ dx$
=$\int\frac{\sqrt{2}}{x-\sqrt{2}}-\frac{\sqrt{2}}{x+\sqrt{2}}\ dx$
=$\sqrt{2}(ln|x-\sqrt{2}|-ln|x+\sqrt{2}|)$
=$\sqrt{2}ln\left|\frac{x-\sqrt{2}}{x+\sqrt{2}}\right|$
Jadi
$\int\frac{x^4}{x^2-2}\ dx$
=$\frac{x^3}{3}+2x+\sqrt{2}ln\left|\frac{x-\sqrt{2}}{x+\sqrt{2}}\right|+C$
30) $\int\frac{sec\ x.tan\ x}{sec\ x+sec^2\ x}\ dx$
Jawab : misalkan $u=sec\ x$
$du=sec\ x.tan\ x\ dx$
$\int\frac{1}{u+u^2}\ du$
=$\int\frac{1}{u(1+u)}\ du$
=$\int\frac{1}{u}-\frac{1}{u+1}\ du$
=$ln|u|-ln|u+1|+C$
=$ln|sec\ x|-ln|sec\ x+1|+C$
=$ln\left|\frac{sec\ x}{sec\ x+1}\right|+C$
31) $\int\frac{x}{(x^2+2x+2)^2}\ dx$
Jawab :
$\int\frac{x}{(x^2+2x+2)^2}\ dx$
=$\int\frac{x+1-1}{(x^2+2x+2)^2}\ dx$
=$\int\frac{x+1}{(x^2+2x+2)^2}\ dx -\int\frac{1}{(x^2+2x+2)^2}\ dx$
Obs :$\int\frac{x+1}{(x^2+2x+2)^2}\ dx$
misalkan :$u=x^2+2x+2$
$\frac{1}{2}\ du=x+1\ dx$
=$\int\frac{1}{2u^2}\ du$
=$\frac{-1}{2u}$
=$\frac{-1}{2x^2+4x+4}$
Obs :$\int\frac{1}{(x^2+2x+2)^2}\ dx$
=$\int\frac{1}{((x+1)^2+1)^2}\ dx$
misalkan :$x+1=tan\ y$
$dx=sec^2\ y\ dy$
=$\int\frac{sec^2\ y}{(tan^2\ y+1)^2}\ dy$
=$\int\frac{1}{sec^2\ y}\ dy$
=$\int cos^2\ y\ dy$
=$\int \frac{cos\ 2y+1}{2}\ dy$
=$\frac{sin\ 2y}{4}+\frac{y}{2}$
=$\frac{sin\ y.cos\ y}{2}+\frac{y}{2}$
=$\frac{\frac{x+1}{x^2+2x+2}}{2}+\frac{tan^{-1}(x+1)}{2}$
=$\frac{x+1}{2x^2+4x+4}+\frac{tan^{-1}(x+1)}{2}$
Jadi,
$\int\frac{x}{(x^2+2x+2)^2}\ dx$
=$\frac{-1}{2x^2+4x+4}-\frac{x+1}{2x^2+4x+4}-\frac{tan^{-1}(x+1)}{2}+C$
=$\frac{-x-2}{2x^2+4x+4}-\frac{tan^{-1}(x+1)}{2}+C$
32) $\int \frac{x^{\frac{1}{3}}}{x^{\frac{1}{2}}+x^{\frac{1}{4}}}\ dx$
Jawab : misalkan $u=x^{\frac{1}{12}}$
$du=\frac{1}{12x^{\frac{11}{12}}}\ dx$
$dx=12x^{\frac{11}{12}}\ du=12u^11\ du$
$\int \frac{u^4}{u^6+u^3}.12u^11\ du$
=$\int \frac{12u^12}{u^3+1}\ du$
Perhatikan bahwa
$u^12=(u^3+1)(u^9-u^6+u^3-1)+1$
$\int \frac{12u^12}{u^3+1}\ du$
=$\int u^9-u^6+u^3-1+\frac{1}{u^3+1}\ du$
=$\frac{u^{10}}{10}-\frac{u^7}{7}+\frac{u^4}{4}-u+\int\frac{1}{u^3+1}\ du$
Obs : $\int\frac{1}{u^3+1}\ du$
=$\int\frac{1}{(u+1)(u^2-u+1)}\ du$
=$\int\frac{1}{3(u+1)}-\frac{u-2}{3(u^2-u+1)}\ du$
=$\frac{1}{3}ln|u+1|-\frac{1}{3}\int\frac{u-2}{u^2-u+1}\ du$
Obs : $\int\frac{u-2}{u^2-u+1}\ du$
=$\frac{1}{2}\int\frac{2u-1-3}{u^2-u+1}\ du$
=$\frac{1}{2}\int\frac{2u-1}{u^2-u+1}\ du -\frac{3}{2}\int\frac{1}{\left(u-\frac{1}{2}\right)^2+\frac{3}{4}}\ du$
=$\frac{1}{2}ln|u^2-u+1|-\frac{3}{2}\int\frac{1}{\left(u-\frac{1}{2}\right)^2+\frac{3}{4}}\ du$
Obs : $\int\frac{1}{\left(u-\frac{1}{2}\right)^2+\frac{3}{4}}\ du$
misalkan $u-\frac{1}{2}=\frac{\sqrt{3}}{2}tan\ v$
$tan\ v=\frac{2u-1}{\sqrt{3}}$
$sin\ v=\frac{2u-1}{(2u-1)^2+3}$
$=\frac{2u-1}{4u^2-4u+4}$
$du=\frac{\sqrt{3}}{2}sec^2\ v\ dv$
=$\int\frac{1}{\frac{3}{4}sec^2\ v}.\frac{\sqrt{3}}{2}sec^2\ v\ dv$
=$\int\frac{2\sqrt{3}}{3}\ dv$
=$\frac{2\sqrt{3}}{3}v$
=$\frac{2\sqrt{3}}{3}tan^{-1}\left(\frac{2u\sqrt{3}}{3}-\frac{\sqrt{3}}{3}\right)$
Jadi,
$\int \frac{x^{\frac{1}{3}}}{x^{\frac{1}{2}}+x^{\frac{1}{4}}}\ dx$
=$\begin{align}&\frac{x^{\frac{10}{12}}}{10}-\frac{x^{\frac{7}{12}}}{7}+\frac{x^{\frac{4}{12}}}{4}-x^{\frac{1}{12}}+\frac{1}{3}ln|x^{\frac{1}{12}}+1|\\ &-\frac{1}{3}\left[\frac{1}{2}ln|x^{\frac{2}{12}}-x^{\frac{1}{12}}+1|-\frac{3}{2}\left(\frac{2\sqrt{3}}{3}tan^{-1}\left(\frac{2x^{\frac{1}{12}}\sqrt{3}}{3}-\frac{\sqrt{3}}{3}\right)\right)\right]+C\end{align}$
=$\begin{align}&\frac{x^{\frac{10}{12}}}{10}-\frac{x^{\frac{7}{12}}}{7}+\frac{x^{\frac{4}{12}}}{4}-x^{\frac{1}{12}}+\frac{1}{3}ln|x^{\frac{1}{12}}+1|\\ &-\frac{1}{6}ln|x^{\frac{2}{12}}-x^{\frac{1}{12}}+1|-\frac{\sqrt{3}}{3}tan^{-1}\left(\frac{2x^{\frac{1}{12}}\sqrt{3}}{3}-\frac{\sqrt{3}}{3}\right)+C\end{align}$
33) $\int\frac{1}{1+cos\ 2\theta}\ d\theta$
Jawab :
$\int\frac{1}{1+cos\ 2\theta}\ d\theta$
=$\int\frac{1}{2cos^2\ \theta}\ d\theta$
=$\frac{1}{2}\int sec^2\ \theta\ d\theta$
=$\frac{1}{2}tan\ \theta+C$
34) $\int\frac{sec\ x}{tan\ x}\ dx$
Jawab :
$\int\frac{sec\ x}{tan\ x}\ dx$
=$\int\frac{1}{sin\ x}\ dx$
=$\int csc\ x\ dx$
=$-ln|csc\ x+cot\ x|+C$
35) $\int sec^3\ x.tan^3\ x\ dx$
Jawab :
$\int sec^3\ x.tan^3\ x\ dx$
=$\int sec^3\ x.(sec^2\ x-1).tan\ x\ dx$
misalkan $u=sec\ x$
$du=sec\ x.tan\ x\ dx$
=$\int u^2(u^2-1)\ du$
=$\int u^4-u^2\ du$
=$\frac{u^5}{5}-\frac{u^3}{3}+C$
=$\frac{sec^5\ x}{5}-\frac{sec^3\ x}{3}+C$
36) $\int x^2tan^{-1}\ x\ dx$
Jawab : misalkan $\begin{align*}u&=tan^{-1}\ x\\
du&=\frac{1}{x^2+1}\ dx\\
dv&=x^2\ dx\\
v&=\frac{x^3}{3}\end{align*}$
$\int u\ dv=uv-\int v\ du$
$\int x^2tan^{-1}\ x\ dx$
=$tan^{-1}\ x.\frac{x^3}{3}-\int \frac{x^3}{3}\frac{1}{x^2+1}\ dx$
=$tan^{-1}\ x.\frac{x^3}{3}-\frac{1}{3}\int \frac{x^3}{(x^2+1)}\ dx$
=$tan^{-1}\ x.\frac{x^3}{3}-\frac{1}{3}\int x+\frac{-x}{(x^2+1)}\ dx$
=$tan^{-1}\ x.\frac{x^3}{3}-\frac{x^2}{6}-\frac{1}{3}\int\frac{-x}{(x^2+1)}\ dx$
Obs : $\int\frac{-x}{(x^2+1)}\ dx$
misalkan $y=x^2+1$
$dy=2x\ dx$
=$\int\frac{-1}{2y}\ dy$
=$\frac{-1}{2}ln|y|$
=$\frac{-1}{2}ln|x^2+1|$ Jadi,
$\int x^2tan^{-1}\ x\ dx$
=$tan^{-1}\ x.\frac{x^3}{3}-\frac{x^2}{6}-\frac{1}{3}\left(\frac{-1}{2}\right)ln|x^2+1|+C$
=$tan^{-1}\ x.\frac{x^3}{3}-\frac{x^2}{6}+\frac{1}{6}ln|x^2+1|+C$
37) $\int x(ln\ x)^3\ dx$
Jawab : misalkan $\begin{align*}u&=ln\ x\\
x&=e^{u}\\
dx&=e^{u}\ du\end{align*}$
$\int x(ln\ x)^3\ dx$
=$\int e^{2u}u^3\ du$
$\begin{align*}y&=u^3\, \,&z''''=e^{2u}\\
y'&=3u^2\, \,&z'''=\frac{e^{2u}}{2}\\
y''&=6u\, \,&z''=\frac{e^{2u}}{4}\\
y'''&=6\, \,&z'=\frac{e^{2u}}{8}\\
y''''&=0\, \,&z=\frac{e^{2u}}{16}
\end{align*}$
$\int e^{2u}u^3\ du$
=$y.z'''-y'.z''+y''.z'-y'''.z$
=$\frac{u^3.e^{2u}}{2}-\frac{3u^2.e^{2u}}{4}+\frac{3u.e^{2u}}{4}-\frac{3.e^{2u}}{8}$
Jadi, $\int x(ln\ x)^3\ dx$
=$\frac{x^{2}(ln\ x)^3}{2}-\frac{3x^{2}(ln\ x)^2}{4}+\frac{3x^{2}ln\ x}{4}-\frac{3x^{2}}{8}+C$
38) $\int \frac{1}{x\sqrt{1+x^2}}\ dx$
Jawab : misalkan $x=tan\ \theta$
$dx=sec^2\ \theta\ d\theta$
$\int \frac{1}{x\sqrt{1+x^2}}\ dx$
=$\int \frac{sec\ \theta}{tan\ \theta}\ d\theta$
=$\int \frac{1}{sin\ \theta}\ d\theta$
=$-csc\ \theta.cot\ \theta+C$
=$-\frac{\sqrt{1+x^2}}{x}.\frac{1}{x}+C$
=$-\frac{\sqrt{1+x^2}}{x^2} +C$
39) $\int e^{x}\sqrt{1+e^{2x}}\ dx$
Jawab : misalkan $y=e^{x}$
$x=ln\ y$
$dx=\frac{1}{y}\ dy$
$\int e^{x}\sqrt{1+e^{2x}}\ dx$
=$\int \sqrt{1+y^2}\ dy$
=$\frac{1}{3y}(1+y^2)\sqrt{1+y^2}+C$
=$\frac{1}{3e^{x}}(1+e^{2x})\sqrt{1+e^{2x}}+C$
40) $\int \frac{x}{\sqrt{4x-x^2}}\ dx$
Jawab :
$\int \frac{x}{\sqrt{4x-x^2}}\ dx$
=$\int \frac{x-2}{\sqrt{4x-x^2}}\ dx+\int \frac{2}{\sqrt{4x-x^2}}\ dx$
Obs : $\int \frac{x-2}{\sqrt{4x-x^2}}\ dx$
misalkan : $u=4x-x^2$
$du=4-2x\ dx=-2(x-2)\ dx$
=$\int \frac{-1}{2\sqrt{u}}\ du$
=$-\sqrt{u}$
=$-\sqrt{4x-x^2}$
Obs : $\int \frac{2}{\sqrt{4x-x^2}}\ dx$
=$\frac{4\sqrt{4x-x^2}}{4-2x}$
=$\frac{2\sqrt{4x-x^2}}{2-x}$
Jadi, $\int \frac{x}{\sqrt{4x-x^2}}\ dx$
=$-\sqrt{4x-x^2}+\frac{2\sqrt{4x-x^2}}{2-x}+C$
41) $\int \frac{1}{x^3\sqrt{x^2-9}}\ dx$
Jawab : Misalkan $x=3sec\ \theta$
$dx=3sec\ \theta.tan\ \theta\ d\theta$
$\int \frac{3sec\ \theta.tan\ \theta}{81sec^3\ \theta.tan\ \theta}\ d\theta$
=$\int \frac{1}{27sec^2\ \theta}\ d\theta$
=$\frac{-1}{27sec\ theta}.\frac{1}{sec\ \theta.tan\ \theta}+C$
=$\frac{-1}{27sec^2\ theta.tan\ \theta}+C$
=$\frac{-1}{27sec^2\ theta.tan\ \theta}+C$
=$\frac{-1}{3x^2.\frac{\sqrt{x^2-9}}{9}}+C$
=$\frac{-3}{x^2\sqrt{x^2-9}}+C$
42) $\int \frac{x}{(7x+1)^{17}}\ dx$
Jawab : Misalkan $\begin{align*}u&=x\\
du&=dx\\
dv&=(7x+1)^{-17}\ dx\\
v&=\frac{(7x+1)^{-16}}{-16.7}\\
&=-\frac{(7x+1)^{-16}}{112}\end{align*}$
Maka dengan menggunakan parsial didapat
$\int \frac{x}{(7x+1)^{17}}\ dx$
=$\frac{-x(7x+1)^{-16}}{112}+\int \frac{(7x+1)^{-16}}{112}\ dx$
=$\frac{-x(7x+1)^{-16}}{112}+\frac{(7x+1)^{-15}}{112.(-15).7}+C$
=$\frac{-x}{112(7x+1)^{16}}-\frac{1}{3920(7x+1)^{15}}+C$
43) $\int \frac{4x^2+x+1}{4x^3+x}\ dx$
Jawab : Perhatikan bahwa
$\int \frac{4x^2+x+1}{4x^3+x}\ dx$
=$\int \frac{1}{4x^2+1}\ dx+\int \frac{1}{x}\ dx$
Langsung saja pake formula $\int \frac{1}{(ax)^2+1}\ dx=\frac{tan^{-1}\ ax}{a}$
Jadi hasilnya adalah
=$\frac{tan^{-1}\ 2x}{2}+ln|x|+C$
44) $\int \frac{4x^3-x+1}{x^3+1}\ dx$
Jawab : $\int \frac{4x^3-x+1}{x^3+1}\ dx$
=$\int \frac{4(x^3+1)-x-3}{x^3+1}\ dx$
=$\int 4-\frac{x+3}{x^3+1}\ dx$
=$4x-\int \frac{x+3}{x^3+1}\ dx$
=$4x-\frac{1}{3}\int \frac{7-x}{x^2-x+1}+\frac{2}{x+1}\ dx$
=$4x-\frac{1}{3}\int \frac{\frac{-1}{2}(2x-1)}{x^2-x+1}+\frac{13}{2(x^2-x+1)}+\frac{2}{x+1}\ dx$
=$\begin{align}&4x-\frac{-1}{6}ln|x^2-x+1|+\frac{13}{6}\int \frac{1}{x^2-x+1}\ dx\\ &+2ln|x+1|\end{align}$
Obs : $\int \frac{1}{x^2-x+1}\ dx$
=$\int \frac{1}{\left(x-\frac{1}{2}\right)^2+\frac{3}{4}}\ dx$
misalkan : $x-\frac{1}{2}=\frac{\sqrt{3}}{2}tan\ u$
$dx=\frac{\sqrt{3}}{2}sec^2\ u\ du$
=$\int \frac{\frac{\sqrt{3}}{2}sec^2\ u}{\frac{3}{4}sec^2\ u}\ du$
=$\int \frac{2\sqrt{3}}{3}\ du$
=$\frac{2\sqrt{3}}{3}u$
=$\frac{2\sqrt{3}}{3}tan^{-1}\left(\frac{2x-1}{\sqrt{3}}\right)$ Jadi, $\int \frac{4x^3-x+1}{x^3+1}\ dx$
=$\begin{align}&4x-\frac{-1}{6}ln|x^2-x+1|+\frac{26\sqrt{3}}{18}tan^{-1}\left(\frac{2x-1}{\sqrt{3}}\right)\\ &+2ln|x+1|+C\end{align}$
45) $\int tan^2\ x.sec\ x\ dx$
Jawab : $\int tan^2\ x.sec\ x\ dx$
=$\int sec^3\ x+sec\ x\ dx$
=$\frac{sec\ x.tan\ x}{2}+\frac{1}{2}\int sec\ x\ dx+\int sec\ x\ dx$
=$\frac{sec\ x.tan\ x}{2}+\frac{3}{2}ln|tan\ x+sec\ x|+C$
46) $\int \frac{x^2+2x+2}{(x+1)^3}\ dx$
Jawab : $\int \frac{x^2+2x+2}{x^3+3x^2+3x+1}\ dx$
=$\frac{1}{3}\int \frac{3x^2+6x+6}{x^3+3x^2+3x+1}\ dx$
=$\frac{1}{3}\int \frac{3x^2+6x+3}{x^3+3x^2+3x+1}\ dx+\int \frac{1}{x^3+3x^2+3x+1}$
=$\frac{1}{3}ln|x^3+3x^2+3x+1|+\int \frac{1}{x^3+3x^2+3x+1}$
=$ln|x+1|+\int \frac{1}{(x+1)^3}$
=$ln|x+1|-\frac{1}{2(x+1)^2}+C$
47) $\int \frac{x^4+2x+2}{x^5+x^4}\ dx$
Jawab : $\int \frac{x^4+2x+2}{x^5+x^4}\ dx$
=$\int \frac{1}{x+1}+\frac{2}{x^4}\ dx$
=$ln|x+1|-\frac{2}{3x^3}+C$
48) $\int \frac{8x^2-4x+7}{(x^2+1)(4x+1)}\ dx$
Jawab : $\int \frac{8x^2-4x+7}{(x^2+1)(4x+1)}\ dx$
=$\int \frac{-1}{x^2+1}+\frac{8}{4x+1}\ dx$
=$-tan^{-1}\ x+2ln|4x+1|+C$
49) $\int \frac{3x^5-x^4+2x^3-12x^2-2x+1}{(x^3-1)^2}\ dx$
Jawab : $\int \frac{3x^5-x^4+2x^3-12x^2-2x+1}{(x-1)^{2}(x^2+x+1)^{2}}\ dx$
=$\int \frac{1}{x-1}-\frac{1}{(x-1)^2}+\frac{2x+1}{x^2+x+1}+\frac{4x+2}{(x^2+x+1)^2}\ dx$
=$ln|x-1|+\frac{1}{x-1}+ln|x^2+x+1|-2\frac{1}{x^2+x+1}+C$
=$ln|(x-1)(x^2+x+1)|+\frac{x^2+x+1-2x+2}{(x-1)(x^2+x+1)}+C$
=$ln|x^3-1|+\frac{x^2-x+3}{x^3-1}+C$
50) $\int \frac{x}{x^4+4x^2+8}\ dx$
Jawab : misalkan $u=x^2$
$du=2x\ dx$
$\int \frac{1}{2(u^2+4u+8)}\ du$
=$\frac{1}{2}\int \frac{1}{(u+2)^2+4}\ du$
Substitusi $u+2=2tan\ y$
$du=2sec^2\ y\ dy$
=$\frac{1}{2}\int \frac{2sec^2\ y}{4sec^2\ y}\ dy$
=$\frac{1}{4}y+C$
=$\frac{1}{4}tan^{-1}\left(\frac{x^2+2}{2}\right)+C$
Alhamdulillah pembahasannya selesai 😀😁
Tunggu pembahasan soal selanjutnya,,
saya pribadi meminta maaf jika ternyata masih banyak kekurangan dalam pembahasan ini. Apabila ada kritik dan masukan tuliskan di kolom komentar. Semoga pembahasannya bisa bermanfaat buat kalian semua...
Terima kasih telah berkunjung ke blog saya....🙏
Salam math lover";💕👌
Assalamu'alaikum Wr.Wb.
Nama saya Hasbiansyah Cahyadi
Disini saya akan melanjutkan pembahasan soal integral sebelumnya silahkan kalian untuk menyimak pembahasannya...
Berikut adalah 100 soal integral yang akan saya bahas
Pembahasan ini akan saya jadikan 4 bagian, untuk soal ini adalah problem no 26-50
 |
| Soal Integral Aljabar dan Trigonometri |
Ini adalah pembahasan part ke 2, yaitu soal no 26-50
26) $\int\frac{x^2}{\sqrt{4-x^2}}\ dx$
Jawab : Misalkan $x=2sin\ y$
$dx=2cos\ y\ dy$
$\int\frac{x^2}{\sqrt{4-x^2}}\ dx$
=$\int\frac{4sin^2\ y}{2cos\ y}.2cos\ y\ dy$
=$\int 4sin^2\ y\ dy$
=$\int 2(1-cos\ 2y)\ dy$
=$\int 2\ dy-\int cos\ 2y\ dy$
=$2y-\frac{sin\ 2y}{2}+C$
=$2y-sin\ y.cos\ y+C$
=$2sin^{-1}\left(\frac{x}{2}\right)-\frac{x}{2}.\left(\frac{\sqrt{4-x^2}}{2}\right)+C$
=$2sin^{-1}\left(\frac{x}{2}\right)-\frac{x\sqrt{4-x^2}}{4}+C$
27) $\int\sqrt{2x-x^2}\ dx$
Jawab :
$\int \sqrt{2x-x^2}\ dx$
=$\int \sqrt{1-(1-x)^2}\ dx$
misalkan : $1-x=sin\ y$
$-dx=cos\ y\ dy$
=$\int\sqrt{1-sin^2\ y}.(-cos\ y)\ dy$
=$\int -cos^2\ y\ dy$
=$\int -\frac{cos\ 2y+1}{2}\ dy$
=$-\int\frac{cos\ 2y}{2}\ dy-\int\frac{1}{2}\ dy$
=$-\frac{sin\ 2y}{4}-\frac{y}{2}+C$
=$-\frac{sin\ y.cos\ y}{2}-\frac{y}{2}+C$
=$-\frac{(1-x)\sqrt{2x-x^2}}{2}-\frac{sin^{-1}(1-x)}{2}+C$
28) $\int\frac{4x-2}{x^3-x}\ dx$
Jawab :
$\int\frac{4x-2}{x(x-1)(x+1)}\ dx$
$\int\left(\frac{2}{x}+\frac{1}{x-1}+\frac{-3}{x+1}\right)\ dx$
=$2ln|x|+ln|x-1|-3ln|x+1|+C$
=$ln\left|\frac{x^2(x-1)}{(x+1)^3}\right|+C$
29) $\int\frac{x^4}{x^2-2}\ dx$
Jawab :
Perhatikan bahwa
$x^4=(x^2-2)^2+4x^2-4=(x^2-2)^2+4(x^2-2)+4$
$\int\frac{x^4}{x^2-2}\ dx$
=$\int (x^2-2)+4+\frac{4}{x^2-2}\ dx$
=$\int x^2+2\ dx+\int\frac{4}{x^2-2}\ dx$
Obs : $\int\frac{4}{x^2-2}\ dx$
=$\int\frac{4}{(x-\sqrt{2})(x+\sqrt{2})}\ dx$
=$\int\frac{\sqrt{2}}{x-\sqrt{2}}-\frac{\sqrt{2}}{x+\sqrt{2}}\ dx$
=$\sqrt{2}(ln|x-\sqrt{2}|-ln|x+\sqrt{2}|)$
=$\sqrt{2}ln\left|\frac{x-\sqrt{2}}{x+\sqrt{2}}\right|$
Jadi
$\int\frac{x^4}{x^2-2}\ dx$
=$\frac{x^3}{3}+2x+\sqrt{2}ln\left|\frac{x-\sqrt{2}}{x+\sqrt{2}}\right|+C$
30) $\int\frac{sec\ x.tan\ x}{sec\ x+sec^2\ x}\ dx$
Jawab : misalkan $u=sec\ x$
$du=sec\ x.tan\ x\ dx$
$\int\frac{1}{u+u^2}\ du$
=$\int\frac{1}{u(1+u)}\ du$
=$\int\frac{1}{u}-\frac{1}{u+1}\ du$
=$ln|u|-ln|u+1|+C$
=$ln|sec\ x|-ln|sec\ x+1|+C$
=$ln\left|\frac{sec\ x}{sec\ x+1}\right|+C$
31) $\int\frac{x}{(x^2+2x+2)^2}\ dx$
Jawab :
$\int\frac{x}{(x^2+2x+2)^2}\ dx$
=$\int\frac{x+1-1}{(x^2+2x+2)^2}\ dx$
=$\int\frac{x+1}{(x^2+2x+2)^2}\ dx -\int\frac{1}{(x^2+2x+2)^2}\ dx$
Obs :$\int\frac{x+1}{(x^2+2x+2)^2}\ dx$
misalkan :$u=x^2+2x+2$
$\frac{1}{2}\ du=x+1\ dx$
=$\int\frac{1}{2u^2}\ du$
=$\frac{-1}{2u}$
=$\frac{-1}{2x^2+4x+4}$
Obs :$\int\frac{1}{(x^2+2x+2)^2}\ dx$
=$\int\frac{1}{((x+1)^2+1)^2}\ dx$
misalkan :$x+1=tan\ y$
$dx=sec^2\ y\ dy$
=$\int\frac{sec^2\ y}{(tan^2\ y+1)^2}\ dy$
=$\int\frac{1}{sec^2\ y}\ dy$
=$\int cos^2\ y\ dy$
=$\int \frac{cos\ 2y+1}{2}\ dy$
=$\frac{sin\ 2y}{4}+\frac{y}{2}$
=$\frac{sin\ y.cos\ y}{2}+\frac{y}{2}$
=$\frac{\frac{x+1}{x^2+2x+2}}{2}+\frac{tan^{-1}(x+1)}{2}$
=$\frac{x+1}{2x^2+4x+4}+\frac{tan^{-1}(x+1)}{2}$
Jadi,
$\int\frac{x}{(x^2+2x+2)^2}\ dx$
=$\frac{-1}{2x^2+4x+4}-\frac{x+1}{2x^2+4x+4}-\frac{tan^{-1}(x+1)}{2}+C$
=$\frac{-x-2}{2x^2+4x+4}-\frac{tan^{-1}(x+1)}{2}+C$
32) $\int \frac{x^{\frac{1}{3}}}{x^{\frac{1}{2}}+x^{\frac{1}{4}}}\ dx$
Jawab : misalkan $u=x^{\frac{1}{12}}$
$du=\frac{1}{12x^{\frac{11}{12}}}\ dx$
$dx=12x^{\frac{11}{12}}\ du=12u^11\ du$
$\int \frac{u^4}{u^6+u^3}.12u^11\ du$
=$\int \frac{12u^12}{u^3+1}\ du$
Perhatikan bahwa
$u^12=(u^3+1)(u^9-u^6+u^3-1)+1$
$\int \frac{12u^12}{u^3+1}\ du$
=$\int u^9-u^6+u^3-1+\frac{1}{u^3+1}\ du$
=$\frac{u^{10}}{10}-\frac{u^7}{7}+\frac{u^4}{4}-u+\int\frac{1}{u^3+1}\ du$
Obs : $\int\frac{1}{u^3+1}\ du$
=$\int\frac{1}{(u+1)(u^2-u+1)}\ du$
=$\int\frac{1}{3(u+1)}-\frac{u-2}{3(u^2-u+1)}\ du$
=$\frac{1}{3}ln|u+1|-\frac{1}{3}\int\frac{u-2}{u^2-u+1}\ du$
Obs : $\int\frac{u-2}{u^2-u+1}\ du$
=$\frac{1}{2}\int\frac{2u-1-3}{u^2-u+1}\ du$
=$\frac{1}{2}\int\frac{2u-1}{u^2-u+1}\ du -\frac{3}{2}\int\frac{1}{\left(u-\frac{1}{2}\right)^2+\frac{3}{4}}\ du$
=$\frac{1}{2}ln|u^2-u+1|-\frac{3}{2}\int\frac{1}{\left(u-\frac{1}{2}\right)^2+\frac{3}{4}}\ du$
Obs : $\int\frac{1}{\left(u-\frac{1}{2}\right)^2+\frac{3}{4}}\ du$
misalkan $u-\frac{1}{2}=\frac{\sqrt{3}}{2}tan\ v$
$tan\ v=\frac{2u-1}{\sqrt{3}}$
$sin\ v=\frac{2u-1}{(2u-1)^2+3}$
$=\frac{2u-1}{4u^2-4u+4}$
$du=\frac{\sqrt{3}}{2}sec^2\ v\ dv$
=$\int\frac{1}{\frac{3}{4}sec^2\ v}.\frac{\sqrt{3}}{2}sec^2\ v\ dv$
=$\int\frac{2\sqrt{3}}{3}\ dv$
=$\frac{2\sqrt{3}}{3}v$
=$\frac{2\sqrt{3}}{3}tan^{-1}\left(\frac{2u\sqrt{3}}{3}-\frac{\sqrt{3}}{3}\right)$
Jadi,
$\int \frac{x^{\frac{1}{3}}}{x^{\frac{1}{2}}+x^{\frac{1}{4}}}\ dx$
=$\begin{align}&\frac{x^{\frac{10}{12}}}{10}-\frac{x^{\frac{7}{12}}}{7}+\frac{x^{\frac{4}{12}}}{4}-x^{\frac{1}{12}}+\frac{1}{3}ln|x^{\frac{1}{12}}+1|\\ &-\frac{1}{3}\left[\frac{1}{2}ln|x^{\frac{2}{12}}-x^{\frac{1}{12}}+1|-\frac{3}{2}\left(\frac{2\sqrt{3}}{3}tan^{-1}\left(\frac{2x^{\frac{1}{12}}\sqrt{3}}{3}-\frac{\sqrt{3}}{3}\right)\right)\right]+C\end{align}$
=$\begin{align}&\frac{x^{\frac{10}{12}}}{10}-\frac{x^{\frac{7}{12}}}{7}+\frac{x^{\frac{4}{12}}}{4}-x^{\frac{1}{12}}+\frac{1}{3}ln|x^{\frac{1}{12}}+1|\\ &-\frac{1}{6}ln|x^{\frac{2}{12}}-x^{\frac{1}{12}}+1|-\frac{\sqrt{3}}{3}tan^{-1}\left(\frac{2x^{\frac{1}{12}}\sqrt{3}}{3}-\frac{\sqrt{3}}{3}\right)+C\end{align}$
33) $\int\frac{1}{1+cos\ 2\theta}\ d\theta$
Jawab :
$\int\frac{1}{1+cos\ 2\theta}\ d\theta$
=$\int\frac{1}{2cos^2\ \theta}\ d\theta$
=$\frac{1}{2}\int sec^2\ \theta\ d\theta$
=$\frac{1}{2}tan\ \theta+C$
34) $\int\frac{sec\ x}{tan\ x}\ dx$
Jawab :
$\int\frac{sec\ x}{tan\ x}\ dx$
=$\int\frac{1}{sin\ x}\ dx$
=$\int csc\ x\ dx$
=$-ln|csc\ x+cot\ x|+C$
35) $\int sec^3\ x.tan^3\ x\ dx$
Jawab :
$\int sec^3\ x.tan^3\ x\ dx$
=$\int sec^3\ x.(sec^2\ x-1).tan\ x\ dx$
misalkan $u=sec\ x$
$du=sec\ x.tan\ x\ dx$
=$\int u^2(u^2-1)\ du$
=$\int u^4-u^2\ du$
=$\frac{u^5}{5}-\frac{u^3}{3}+C$
=$\frac{sec^5\ x}{5}-\frac{sec^3\ x}{3}+C$
36) $\int x^2tan^{-1}\ x\ dx$
Jawab : misalkan $\begin{align*}u&=tan^{-1}\ x\\
du&=\frac{1}{x^2+1}\ dx\\
dv&=x^2\ dx\\
v&=\frac{x^3}{3}\end{align*}$
$\int u\ dv=uv-\int v\ du$
$\int x^2tan^{-1}\ x\ dx$
=$tan^{-1}\ x.\frac{x^3}{3}-\int \frac{x^3}{3}\frac{1}{x^2+1}\ dx$
=$tan^{-1}\ x.\frac{x^3}{3}-\frac{1}{3}\int \frac{x^3}{(x^2+1)}\ dx$
=$tan^{-1}\ x.\frac{x^3}{3}-\frac{1}{3}\int x+\frac{-x}{(x^2+1)}\ dx$
=$tan^{-1}\ x.\frac{x^3}{3}-\frac{x^2}{6}-\frac{1}{3}\int\frac{-x}{(x^2+1)}\ dx$
Obs : $\int\frac{-x}{(x^2+1)}\ dx$
misalkan $y=x^2+1$
$dy=2x\ dx$
=$\int\frac{-1}{2y}\ dy$
=$\frac{-1}{2}ln|y|$
=$\frac{-1}{2}ln|x^2+1|$ Jadi,
$\int x^2tan^{-1}\ x\ dx$
=$tan^{-1}\ x.\frac{x^3}{3}-\frac{x^2}{6}-\frac{1}{3}\left(\frac{-1}{2}\right)ln|x^2+1|+C$
=$tan^{-1}\ x.\frac{x^3}{3}-\frac{x^2}{6}+\frac{1}{6}ln|x^2+1|+C$
37) $\int x(ln\ x)^3\ dx$
Jawab : misalkan $\begin{align*}u&=ln\ x\\
x&=e^{u}\\
dx&=e^{u}\ du\end{align*}$
$\int x(ln\ x)^3\ dx$
=$\int e^{2u}u^3\ du$
$\begin{align*}y&=u^3\, \,&z''''=e^{2u}\\
y'&=3u^2\, \,&z'''=\frac{e^{2u}}{2}\\
y''&=6u\, \,&z''=\frac{e^{2u}}{4}\\
y'''&=6\, \,&z'=\frac{e^{2u}}{8}\\
y''''&=0\, \,&z=\frac{e^{2u}}{16}
\end{align*}$
$\int e^{2u}u^3\ du$
=$y.z'''-y'.z''+y''.z'-y'''.z$
=$\frac{u^3.e^{2u}}{2}-\frac{3u^2.e^{2u}}{4}+\frac{3u.e^{2u}}{4}-\frac{3.e^{2u}}{8}$
Jadi, $\int x(ln\ x)^3\ dx$
=$\frac{x^{2}(ln\ x)^3}{2}-\frac{3x^{2}(ln\ x)^2}{4}+\frac{3x^{2}ln\ x}{4}-\frac{3x^{2}}{8}+C$
38) $\int \frac{1}{x\sqrt{1+x^2}}\ dx$
Jawab : misalkan $x=tan\ \theta$
$dx=sec^2\ \theta\ d\theta$
$\int \frac{1}{x\sqrt{1+x^2}}\ dx$
=$\int \frac{sec\ \theta}{tan\ \theta}\ d\theta$
=$\int \frac{1}{sin\ \theta}\ d\theta$
=$-csc\ \theta.cot\ \theta+C$
=$-\frac{\sqrt{1+x^2}}{x}.\frac{1}{x}+C$
=$-\frac{\sqrt{1+x^2}}{x^2} +C$
39) $\int e^{x}\sqrt{1+e^{2x}}\ dx$
Jawab : misalkan $y=e^{x}$
$x=ln\ y$
$dx=\frac{1}{y}\ dy$
$\int e^{x}\sqrt{1+e^{2x}}\ dx$
=$\int \sqrt{1+y^2}\ dy$
=$\frac{1}{3y}(1+y^2)\sqrt{1+y^2}+C$
=$\frac{1}{3e^{x}}(1+e^{2x})\sqrt{1+e^{2x}}+C$
40) $\int \frac{x}{\sqrt{4x-x^2}}\ dx$
Jawab :
$\int \frac{x}{\sqrt{4x-x^2}}\ dx$
=$\int \frac{x-2}{\sqrt{4x-x^2}}\ dx+\int \frac{2}{\sqrt{4x-x^2}}\ dx$
Obs : $\int \frac{x-2}{\sqrt{4x-x^2}}\ dx$
misalkan : $u=4x-x^2$
$du=4-2x\ dx=-2(x-2)\ dx$
=$\int \frac{-1}{2\sqrt{u}}\ du$
=$-\sqrt{u}$
=$-\sqrt{4x-x^2}$
Obs : $\int \frac{2}{\sqrt{4x-x^2}}\ dx$
=$\frac{4\sqrt{4x-x^2}}{4-2x}$
=$\frac{2\sqrt{4x-x^2}}{2-x}$
Jadi, $\int \frac{x}{\sqrt{4x-x^2}}\ dx$
=$-\sqrt{4x-x^2}+\frac{2\sqrt{4x-x^2}}{2-x}+C$
41) $\int \frac{1}{x^3\sqrt{x^2-9}}\ dx$
Jawab : Misalkan $x=3sec\ \theta$
$dx=3sec\ \theta.tan\ \theta\ d\theta$
$\int \frac{3sec\ \theta.tan\ \theta}{81sec^3\ \theta.tan\ \theta}\ d\theta$
=$\int \frac{1}{27sec^2\ \theta}\ d\theta$
=$\frac{-1}{27sec\ theta}.\frac{1}{sec\ \theta.tan\ \theta}+C$
=$\frac{-1}{27sec^2\ theta.tan\ \theta}+C$
=$\frac{-1}{27sec^2\ theta.tan\ \theta}+C$
=$\frac{-1}{3x^2.\frac{\sqrt{x^2-9}}{9}}+C$
=$\frac{-3}{x^2\sqrt{x^2-9}}+C$
42) $\int \frac{x}{(7x+1)^{17}}\ dx$
Jawab : Misalkan $\begin{align*}u&=x\\
du&=dx\\
dv&=(7x+1)^{-17}\ dx\\
v&=\frac{(7x+1)^{-16}}{-16.7}\\
&=-\frac{(7x+1)^{-16}}{112}\end{align*}$
Maka dengan menggunakan parsial didapat
$\int \frac{x}{(7x+1)^{17}}\ dx$
=$\frac{-x(7x+1)^{-16}}{112}+\int \frac{(7x+1)^{-16}}{112}\ dx$
=$\frac{-x(7x+1)^{-16}}{112}+\frac{(7x+1)^{-15}}{112.(-15).7}+C$
=$\frac{-x}{112(7x+1)^{16}}-\frac{1}{3920(7x+1)^{15}}+C$
43) $\int \frac{4x^2+x+1}{4x^3+x}\ dx$
Jawab : Perhatikan bahwa
$\int \frac{4x^2+x+1}{4x^3+x}\ dx$
=$\int \frac{1}{4x^2+1}\ dx+\int \frac{1}{x}\ dx$
Langsung saja pake formula $\int \frac{1}{(ax)^2+1}\ dx=\frac{tan^{-1}\ ax}{a}$
Jadi hasilnya adalah
=$\frac{tan^{-1}\ 2x}{2}+ln|x|+C$
44) $\int \frac{4x^3-x+1}{x^3+1}\ dx$
Jawab : $\int \frac{4x^3-x+1}{x^3+1}\ dx$
=$\int \frac{4(x^3+1)-x-3}{x^3+1}\ dx$
=$\int 4-\frac{x+3}{x^3+1}\ dx$
=$4x-\int \frac{x+3}{x^3+1}\ dx$
=$4x-\frac{1}{3}\int \frac{7-x}{x^2-x+1}+\frac{2}{x+1}\ dx$
=$4x-\frac{1}{3}\int \frac{\frac{-1}{2}(2x-1)}{x^2-x+1}+\frac{13}{2(x^2-x+1)}+\frac{2}{x+1}\ dx$
=$\begin{align}&4x-\frac{-1}{6}ln|x^2-x+1|+\frac{13}{6}\int \frac{1}{x^2-x+1}\ dx\\ &+2ln|x+1|\end{align}$
Obs : $\int \frac{1}{x^2-x+1}\ dx$
=$\int \frac{1}{\left(x-\frac{1}{2}\right)^2+\frac{3}{4}}\ dx$
misalkan : $x-\frac{1}{2}=\frac{\sqrt{3}}{2}tan\ u$
$dx=\frac{\sqrt{3}}{2}sec^2\ u\ du$
=$\int \frac{\frac{\sqrt{3}}{2}sec^2\ u}{\frac{3}{4}sec^2\ u}\ du$
=$\int \frac{2\sqrt{3}}{3}\ du$
=$\frac{2\sqrt{3}}{3}u$
=$\frac{2\sqrt{3}}{3}tan^{-1}\left(\frac{2x-1}{\sqrt{3}}\right)$ Jadi, $\int \frac{4x^3-x+1}{x^3+1}\ dx$
=$\begin{align}&4x-\frac{-1}{6}ln|x^2-x+1|+\frac{26\sqrt{3}}{18}tan^{-1}\left(\frac{2x-1}{\sqrt{3}}\right)\\ &+2ln|x+1|+C\end{align}$
45) $\int tan^2\ x.sec\ x\ dx$
Jawab : $\int tan^2\ x.sec\ x\ dx$
=$\int sec^3\ x+sec\ x\ dx$
=$\frac{sec\ x.tan\ x}{2}+\frac{1}{2}\int sec\ x\ dx+\int sec\ x\ dx$
=$\frac{sec\ x.tan\ x}{2}+\frac{3}{2}ln|tan\ x+sec\ x|+C$
46) $\int \frac{x^2+2x+2}{(x+1)^3}\ dx$
Jawab : $\int \frac{x^2+2x+2}{x^3+3x^2+3x+1}\ dx$
=$\frac{1}{3}\int \frac{3x^2+6x+6}{x^3+3x^2+3x+1}\ dx$
=$\frac{1}{3}\int \frac{3x^2+6x+3}{x^3+3x^2+3x+1}\ dx+\int \frac{1}{x^3+3x^2+3x+1}$
=$\frac{1}{3}ln|x^3+3x^2+3x+1|+\int \frac{1}{x^3+3x^2+3x+1}$
=$ln|x+1|+\int \frac{1}{(x+1)^3}$
=$ln|x+1|-\frac{1}{2(x+1)^2}+C$
47) $\int \frac{x^4+2x+2}{x^5+x^4}\ dx$
Jawab : $\int \frac{x^4+2x+2}{x^5+x^4}\ dx$
=$\int \frac{1}{x+1}+\frac{2}{x^4}\ dx$
=$ln|x+1|-\frac{2}{3x^3}+C$
48) $\int \frac{8x^2-4x+7}{(x^2+1)(4x+1)}\ dx$
Jawab : $\int \frac{8x^2-4x+7}{(x^2+1)(4x+1)}\ dx$
=$\int \frac{-1}{x^2+1}+\frac{8}{4x+1}\ dx$
=$-tan^{-1}\ x+2ln|4x+1|+C$
49) $\int \frac{3x^5-x^4+2x^3-12x^2-2x+1}{(x^3-1)^2}\ dx$
Jawab : $\int \frac{3x^5-x^4+2x^3-12x^2-2x+1}{(x-1)^{2}(x^2+x+1)^{2}}\ dx$
=$\int \frac{1}{x-1}-\frac{1}{(x-1)^2}+\frac{2x+1}{x^2+x+1}+\frac{4x+2}{(x^2+x+1)^2}\ dx$
=$ln|x-1|+\frac{1}{x-1}+ln|x^2+x+1|-2\frac{1}{x^2+x+1}+C$
=$ln|(x-1)(x^2+x+1)|+\frac{x^2+x+1-2x+2}{(x-1)(x^2+x+1)}+C$
=$ln|x^3-1|+\frac{x^2-x+3}{x^3-1}+C$
50) $\int \frac{x}{x^4+4x^2+8}\ dx$
Jawab : misalkan $u=x^2$
$du=2x\ dx$
$\int \frac{1}{2(u^2+4u+8)}\ du$
=$\frac{1}{2}\int \frac{1}{(u+2)^2+4}\ du$
Substitusi $u+2=2tan\ y$
$du=2sec^2\ y\ dy$
=$\frac{1}{2}\int \frac{2sec^2\ y}{4sec^2\ y}\ dy$
=$\frac{1}{4}y+C$
=$\frac{1}{4}tan^{-1}\left(\frac{x^2+2}{2}\right)+C$
Alhamdulillah pembahasannya selesai 😀😁
Tunggu pembahasan soal selanjutnya,,
saya pribadi meminta maaf jika ternyata masih banyak kekurangan dalam pembahasan ini. Apabila ada kritik dan masukan tuliskan di kolom komentar. Semoga pembahasannya bisa bermanfaat buat kalian semua...
Terima kasih telah berkunjung ke blog saya....🙏
Salam math lover";💕👌
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