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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

Soal Integral Aljabar dan Trigonometri

Ini adalah pembahasan part ke 2, yaitu soal no 26-50

26) x24x2 dx
Jawab : Misalkan x=2sin y
                                dx=2cos y dy
x24x2 dx
=4sin2 y2cos y.2cos y dy
=4sin2 y dy
=2(1cos 2y) dy
=2 dycos 2y dy
=2ysin 2y2+C
=2ysin y.cos y+C
=2sin1(x2)x2.(4x22)+C
=2sin1(x2)x4x24+C
27) 2xx2 dx
Jawab :
2xx2 dx
=1(1x)2 dx
misalkan : 1x=sin y
                    dx=cos y dy
=1sin2 y.(cos y) dy
=cos2 y dy
=cos 2y+12 dy
=cos 2y2 dy12 dy
=sin 2y4y2+C
=sin y.cos y2y2+C
=(1x)2xx22sin1(1x)2+C
28) 4x2x3x dx
Jawab :
4x2x(x1)(x+1) dx
(2x+1x1+3x+1) dx
=2ln|x|+ln|x1|3ln|x+1|+C
=ln|x2(x1)(x+1)3|+C
29) x4x22 dx
Jawab :
Perhatikan bahwa
x4=(x22)2+4x24=(x22)2+4(x22)+4
x4x22 dx
=(x22)+4+4x22 dx
=x2+2 dx+4x22 dx
Obs : 4x22 dx
=4(x2)(x+2) dx
=2x22x+2 dx
=2(ln|x2|ln|x+2|)
=2ln|x2x+2|
Jadi
x4x22 dx
=x33+2x+2ln|x2x+2|+C
30) sec x.tan xsec x+sec2 x dx
Jawab : misalkan u=sec x
                                du=sec x.tan x dx
1u+u2 du
=1u(1+u) du
=1u1u+1 du
=ln|u|ln|u+1|+C
=ln|sec x|ln|sec x+1|+C
=ln|sec xsec x+1|+C
31) x(x2+2x+2)2 dx
Jawab :
x(x2+2x+2)2 dx
=x+11(x2+2x+2)2 dx
=x+1(x2+2x+2)2 dx1(x2+2x+2)2 dx
Obs :x+1(x2+2x+2)2 dx
misalkan :u=x2+2x+2
           12 du=x+1 dx
=12u2 du
=12u
=12x2+4x+4
Obs :1(x2+2x+2)2 dx
=1((x+1)2+1)2 dx
misalkan :x+1=tan y
                   dx=sec2 y dy
=sec2 y(tan2 y+1)2 dy
=1sec2 y dy
=cos2 y dy
=cos 2y+12 dy
=sin 2y4+y2
=sin y.cos y2+y2
=x+1x2+2x+22+tan1(x+1)2
=x+12x2+4x+4+tan1(x+1)2
Jadi,
x(x2+2x+2)2 dx
=12x2+4x+4x+12x2+4x+4tan1(x+1)2+C
=x22x2+4x+4tan1(x+1)2+C
32) x13x12+x14 dx
Jawab : misalkan u=x112
                              du=112x1112 dx
                              dx=12x1112 du=12u11 du
u4u6+u3.12u11 du
=12u12u3+1 du
Perhatikan bahwa
u12=(u3+1)(u9u6+u31)+1
12u12u3+1 du
=u9u6+u31+1u3+1 du
=u1010u77+u44u+1u3+1 du
Obs : 1u3+1 du
=1(u+1)(u2u+1) du
=13(u+1)u23(u2u+1) du
=13ln|u+1|13u2u2u+1 du
Obs : u2u2u+1 du
=122u13u2u+1 du
=122u1u2u+1 du321(u12)2+34 du
=12ln|u2u+1|321(u12)2+34 du
Obs : 1(u12)2+34 du
misalkan u12=32tan v
                 tan v=2u13
                 sin v=2u1(2u1)2+3
                           =2u14u24u+4
                 du=32sec2 v dv
=134sec2 v.32sec2 v dv
=233 dv
=233v
=233tan1(2u3333)
Jadi,
x13x12+x14 dx
=x101210x7127+x4124x112+13ln|x112+1|13[12ln|x212x112+1|32(233tan1(2x1123333))]+C
=x101210x7127+x4124x112+13ln|x112+1|16ln|x212x112+1|33tan1(2x1123333)+C
33) 11+cos 2θ dθ
Jawab :
11+cos 2θ dθ
=12cos2 θ dθ
=12sec2 θ dθ
=12tan θ+C
34) sec xtan x dx
Jawab :
sec xtan x dx
=1sin x dx
=csc x dx
=ln|csc x+cot x|+C
35) sec3 x.tan3 x dx
Jawab :
sec3 x.tan3 x dx
=sec3 x.(sec2 x1).tan x dx
misalkan u=sec x
                du=sec x.tan x dx
=u2(u21) du
=u4u2 du
=u55u33+C
=sec5 x5sec3 x3+C
36) x2tan1 x dx
Jawab : misalkan u=tan1 xdu=1x2+1 dxdv=x2 dxv=x33
u dv=uvv du
x2tan1 x dx
=tan1 x.x33x331x2+1 dx
=tan1 x.x3313x3(x2+1) dx
=tan1 x.x3313x+x(x2+1) dx
=tan1 x.x33x2613x(x2+1) dx
Obs : x(x2+1) dx
misalkan y=x2+1
                  dy=2x dx
=12y dy
=12ln|y|
=12ln|x2+1| Jadi,
x2tan1 x dx
=tan1 x.x33x2613(12)ln|x2+1|+C
=tan1 x.x33x26+16ln|x2+1|+C
37) x(ln x)3 dx
Jawab : misalkan u=ln xx=eudx=eu du
x(ln x)3 dx
=e2uu3 du
y=u3z
\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...


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Salam math lover";💕👌