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) ∫x2√4−x2 dx
Jawab : Misalkan x=2sin y
dx=2cos y dy
∫x2√4−x2 dx
=∫4sin2 y2cos y.2cos y dy
=∫4sin2 y dy
=∫2(1−cos 2y) dy
=∫2 dy−∫cos 2y dy
=2y−sin 2y2+C
=2y−sin y.cos y+C
=2sin−1(x2)−x2.(√4−x22)+C
=2sin−1(x2)−x√4−x24+C
27) ∫√2x−x2 dx
Jawab :
∫√2x−x2 dx
=∫√1−(1−x)2 dx
misalkan : 1−x=sin y
−dx=cos y dy
=∫√1−sin2 y.(−cos y) dy
=∫−cos2 y dy
=∫−cos 2y+12 dy
=−∫cos 2y2 dy−∫12 dy
=−sin 2y4−y2+C
=−sin y.cos y2−y2+C
=−(1−x)√2x−x22−sin−1(1−x)2+C
28) ∫4x−2x3−x dx
Jawab :
∫4x−2x(x−1)(x+1) dx
∫(2x+1x−1+−3x+1) dx
=2ln|x|+ln|x−1|−3ln|x+1|+C
=ln|x2(x−1)(x+1)3|+C
29) ∫x4x2−2 dx
Jawab :
Perhatikan bahwa
x4=(x2−2)2+4x2−4=(x2−2)2+4(x2−2)+4
∫x4x2−2 dx
=∫(x2−2)+4+4x2−2 dx
=∫x2+2 dx+∫4x2−2 dx
Obs : ∫4x2−2 dx
=∫4(x−√2)(x+√2) dx
=∫√2x−√2−√2x+√2 dx
=√2(ln|x−√2|−ln|x+√2|)
=√2ln|x−√2x+√2|
Jadi
∫x4x2−2 dx
=x33+2x+√2ln|x−√2x+√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
=∫1u−1u+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+1−1(x2+2x+2)2 dx
=∫x+1(x2+2x+2)2 dx−∫1(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+tan−1(x+1)2
=x+12x2+4x+4+tan−1(x+1)2
Jadi,
∫x(x2+2x+2)2 dx
=−12x2+4x+4−x+12x2+4x+4−tan−1(x+1)2+C
=−x−22x2+4x+4−tan−1(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)(u9−u6+u3−1)+1
∫12u12u3+1 du
=∫u9−u6+u3−1+1u3+1 du
=u1010−u77+u44−u+∫1u3+1 du
Obs : ∫1u3+1 du
=∫1(u+1)(u2−u+1) du
=∫13(u+1)−u−23(u2−u+1) du
=13ln|u+1|−13∫u−2u2−u+1 du
Obs : ∫u−2u2−u+1 du
=12∫2u−1−3u2−u+1 du
=12∫2u−1u2−u+1 du−32∫1(u−12)2+34 du
=12ln|u2−u+1|−32∫1(u−12)2+34 du
Obs : ∫1(u−12)2+34 du
misalkan u−12=√32tan v
tan v=2u−1√3
sin v=2u−1(2u−1)2+3
=2u−14u2−4u+4
du=√32sec2 v dv
=∫134sec2 v.√32sec2 v dv
=∫2√33 dv
=2√33v
=2√33tan−1(2u√33−√33)
Jadi,
∫x13x12+x14 dx
=x101210−x7127+x4124−x112+13ln|x112+1|−13[12ln|x212−x112+1|−32(2√33tan−1(2x112√33−√33))]+C
=x101210−x7127+x4124−x112+13ln|x112+1|−16ln|x212−x112+1|−√33tan−1(2x112√33−√33)+C
33) ∫11+cos 2θ dθ
Jawab :
∫11+cos 2θ dθ
=∫12cos2 θ dθ
=12∫sec2 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 x−1).tan x dx
misalkan u=sec x
du=sec x.tan x dx
=∫u2(u2−1) du
=∫u4−u2 du
=u55−u33+C
=sec5 x5−sec3 x3+C
36) ∫x2tan−1 x dx
Jawab : misalkan u=tan−1 xdu=1x2+1 dxdv=x2 dxv=x33
∫u dv=uv−∫v du
∫x2tan−1 x dx
=tan−1 x.x33−∫x331x2+1 dx
=tan−1 x.x33−13∫x+−x(x2+1) dx
=tan−1 x.x33−x26−13∫−x(x2+1) dx
Obs : ∫−x(x2+1) dx
misalkan y=x2+1
dy=2x dx
=∫−12y dy
=−12ln|y|
=−12ln|x2+1| Jadi,
∫x2tan−1 x dx
=tan−1 x.x33−x26−13(−12)ln|x2+1|+C
=tan−1 x.x33−x26+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...
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) ∫x2√4−x2 dx
Jawab : Misalkan x=2sin y
dx=2cos y dy
∫x2√4−x2 dx
=∫4sin2 y2cos y.2cos y dy
=∫4sin2 y dy
=∫2(1−cos 2y) dy
=∫2 dy−∫cos 2y dy
=2y−sin 2y2+C
=2y−sin y.cos y+C
=2sin−1(x2)−x2.(√4−x22)+C
=2sin−1(x2)−x√4−x24+C
27) ∫√2x−x2 dx
Jawab :
∫√2x−x2 dx
=∫√1−(1−x)2 dx
misalkan : 1−x=sin y
−dx=cos y dy
=∫√1−sin2 y.(−cos y) dy
=∫−cos2 y dy
=∫−cos 2y+12 dy
=−∫cos 2y2 dy−∫12 dy
=−sin 2y4−y2+C
=−sin y.cos y2−y2+C
=−(1−x)√2x−x22−sin−1(1−x)2+C
28) ∫4x−2x3−x dx
Jawab :
∫4x−2x(x−1)(x+1) dx
∫(2x+1x−1+−3x+1) dx
=2ln|x|+ln|x−1|−3ln|x+1|+C
=ln|x2(x−1)(x+1)3|+C
29) ∫x4x2−2 dx
Jawab :
Perhatikan bahwa
x4=(x2−2)2+4x2−4=(x2−2)2+4(x2−2)+4
∫x4x2−2 dx
=∫(x2−2)+4+4x2−2 dx
=∫x2+2 dx+∫4x2−2 dx
Obs : ∫4x2−2 dx
=∫4(x−√2)(x+√2) dx
=∫√2x−√2−√2x+√2 dx
=√2(ln|x−√2|−ln|x+√2|)
=√2ln|x−√2x+√2|
Jadi
∫x4x2−2 dx
=x33+2x+√2ln|x−√2x+√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
=∫1u−1u+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+1−1(x2+2x+2)2 dx
=∫x+1(x2+2x+2)2 dx−∫1(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+tan−1(x+1)2
=x+12x2+4x+4+tan−1(x+1)2
Jadi,
∫x(x2+2x+2)2 dx
=−12x2+4x+4−x+12x2+4x+4−tan−1(x+1)2+C
=−x−22x2+4x+4−tan−1(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)(u9−u6+u3−1)+1
∫12u12u3+1 du
=∫u9−u6+u3−1+1u3+1 du
=u1010−u77+u44−u+∫1u3+1 du
Obs : ∫1u3+1 du
=∫1(u+1)(u2−u+1) du
=∫13(u+1)−u−23(u2−u+1) du
=13ln|u+1|−13∫u−2u2−u+1 du
Obs : ∫u−2u2−u+1 du
=12∫2u−1−3u2−u+1 du
=12∫2u−1u2−u+1 du−32∫1(u−12)2+34 du
=12ln|u2−u+1|−32∫1(u−12)2+34 du
Obs : ∫1(u−12)2+34 du
misalkan u−12=√32tan v
tan v=2u−1√3
sin v=2u−1(2u−1)2+3
=2u−14u2−4u+4
du=√32sec2 v dv
=∫134sec2 v.√32sec2 v dv
=∫2√33 dv
=2√33v
=2√33tan−1(2u√33−√33)
Jadi,
∫x13x12+x14 dx
=x101210−x7127+x4124−x112+13ln|x112+1|−13[12ln|x212−x112+1|−32(2√33tan−1(2x112√33−√33))]+C
=x101210−x7127+x4124−x112+13ln|x112+1|−16ln|x212−x112+1|−√33tan−1(2x112√33−√33)+C
33) ∫11+cos 2θ dθ
Jawab :
∫11+cos 2θ dθ
=∫12cos2 θ dθ
=12∫sec2 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 x−1).tan x dx
misalkan u=sec x
du=sec x.tan x dx
=∫u2(u2−1) du
=∫u4−u2 du
=u55−u33+C
=sec5 x5−sec3 x3+C
36) ∫x2tan−1 x dx
Jawab : misalkan u=tan−1 xdu=1x2+1 dxdv=x2 dxv=x33
∫u dv=uv−∫v du
∫x2tan−1 x dx
=tan−1 x.x33−∫x331x2+1 dx
=tan−1 x.x33−13∫x3(x2+1) dx
=tan−1 x.x33−13∫x+−x(x2+1) dx
=tan−1 x.x33−x26−13∫−x(x2+1) dx
Obs : ∫−x(x2+1) dx
misalkan y=x2+1
dy=2x dx
=∫−12y dy
=−12ln|y|
=−12ln|x2+1| Jadi,
∫x2tan−1 x dx
=tan−1 x.x33−x26−13(−12)ln|x2+1|+C
=tan−1 x.x33−x26+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...
Terima kasih telah berkunjung ke blog saya....🙏
Salam math lover";💕👌
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