31. ∫ 1 x − x 3 / 2 d x \int \frac{1}{\sqrt{x-x^{3/2}}}dx ∫ x − x 3 / 2 1 d x
∫ 1 x − x 3 / 2 d x = ∫ 1 x 1 − x 1 / 2 d x = ∫ x − 1 / 2 1 − x 1 / 2 d x ( u = 1 − x 1 / 2 , d u = − 1 2 x − 1 / 2 d x ) = − 2 ∫ 1 u 1 / 2 d u = − 2 ∫ u − 1 / 2 d u = − 2 ( 2 ) u 1 / 2 = − 4 u = − 4 1 − x + C
\begin{aligned}
&\int \frac{1}{\sqrt{x-x^{3/2}}}dx\\
&=\int \frac{1}{\sqrt{x}\sqrt{1-x^{1/2}}}dx\\
&=\int \frac{x^{-1/2}}{\sqrt{1-x^{1/2}}}dx\\
&(u=1-x^{1/2}, du=-\frac{1}{2}x^{-1/2} dx) \\
&=-2\int \frac{1}{u^{1/2}} du=-2\int u^{-1/2} du\\
&=-2 (2)u^{1/2}=-4\sqrt{u}\\
&=-4\sqrt{1-\sqrt{x}}+C
\end{aligned}
∫ x − x 3 / 2 1 d x = ∫ x 1 − x 1 / 2 1 d x = ∫ 1 − x 1 / 2 x − 1 / 2 d x ( u = 1 − x 1 / 2 , d u = − 2 1 x − 1 / 2 d x ) = − 2 ∫ u 1 / 2 1 d u = − 2 ∫ u − 1 / 2 d u = − 2 ( 2 ) u 1 / 2 = − 4 u = − 4 1 − x + C
32. ∫ 1 x − x 2 d x \int \frac{1}{\sqrt{x-x^2}}dx ∫ x − x 2 1 d x
∫ 1 x − x 2 d x = ∫ 1 x x − 1 − 1 d x = ∫ x − 1 x − 1 − 1 d x ( u = x − 1 − 1 , d u = − x − 2 d x ) = − ∫ x − 1 u x 2 d u
\begin{aligned}
&\int \frac{1}{\sqrt{x-x^2}}dx \\
&=\int \frac{1}{x\sqrt{x^{-1}-1}} dx
=\int \frac{x^{-1}}{\sqrt{x^{-1}-1}} dx\\
&(u=x^{-1}-1, du=-x^{-2}dx)\\
&=-\int \frac{x^{-1}}{\sqrt{u}}x^2du
\end{aligned}
∫ x − x 2 1 d x = ∫ x x − 1 − 1 1 d x = ∫ x − 1 − 1 x − 1 d x ( u = x − 1 − 1 , d u = − x − 2 d x ) = − ∫ u x − 1 x 2 d u
∫ 1 x − x 2 d x = ∫ 1 x 1 − x d x ( u = x , d u = 1 2 x d x ) = ∫ 1 u 1 − u 2 2 u d u = 2 ∫ 1 1 − u 2 d u
\begin{aligned}
&\int \frac{1}{\sqrt{x-x^2}}dx \\
&=\int \frac{1}{\sqrt{x}\sqrt{1-x}} dx \\
&(u=\sqrt{x}, du=\frac{1}{2\sqrt{x}}dx)\\
&=\int \frac{1}{u\sqrt{1-u^2}} 2udu\\
&=2\int \frac{1}{\sqrt{1-u^2}}du \\
\end{aligned}
∫ x − x 2 1 d x = ∫ x 1 − x 1 d x ( u = x , d u = 2 x 1 d x ) = ∫ u 1 − u 2 1 2 u d u = 2 ∫ 1 − u 2 1 d u θ \theta θ = 2 ∫ 1 cos θ cos θ d θ = 2 θ = 2 arcsin u + C = 2 arcsin x + C
=2\int \frac{1}{\cos \theta} \cos \theta d\theta
=2\theta = 2 \arcsin{u}+C\\
=2\arcsin{\sqrt{x}}+C
= 2 ∫ cos θ 1 cos θ d θ = 2 θ = 2 arcsin u + C = 2 arcsin x + C
33. ∫ e 2 l n x d x \int e^{2lnx} dx ∫ e 2 l n x d x
∫ e 2 l n x d x = ∫ e l n x e l n x d x = ∫ ( e l n x ) 2 d x = ∫ x 2 d x o r = ∫ e l n x 2 d x = ∫ x 2 d x = 1 3 x 3 + C
\begin{aligned}
&\int e^{2lnx} dx\\
&=\int e^{lnx}e^{lnx} dx =\int (e^{lnx})^2 dx =\int x^2 dx\\
or&=\int e^{lnx^2} dx =\int x^2 dx\\
&=\frac{1}{3}x^3+C
\end{aligned}
o r ∫ e 2 l n x d x = ∫ e l n x e l n x d x = ∫ ( e l n x ) 2 d x = ∫ x 2 d x = ∫ e l n x 2 d x = ∫ x 2 d x = 3 1 x 3 + C
34. ∫ l n x / s q r t x d x \int lnx/sqrt x dx ∫ l n x / s q r t x d x
∫ ln x x d x ( u = x , d u = 1 2 x d x ) = 2 ∫ l n u 2 d u = 4 ∫ l n u d u = 4 ( u l n ∣ u ∣ − u ) = 4 ( x l n ∣ x ∣ − x ) + C = 2 x l n ( x ) − 4 x + C
\begin{aligned}
&\int \frac{\ln x}{\sqrt x}dx \\
&(u=\sqrt x , du = \frac{1}{2\sqrt x}dx)\\
&=2\int ln u^2 du=4\int ln u du \\
&=4 (uln |u| -u ) \\
&=4 (\sqrt x ln |\sqrt x| - \sqrt x ) + C\\
&=2\sqrt x ln (x) - 4\sqrt x + C\\
\end{aligned}
∫ x ln x d x ( u = x , d u = 2 x 1 d x ) = 2 ∫ l n u 2 d u = 4 ∫ l n u d u = 4 ( u l n ∣ u ∣ − u ) = 4 ( x l n ∣ x ∣ − x ) + C = 2 x l n ( x ) − 4 x + C
∫ l n x d x = ( l n x ) x − ∫ 1 / x ∗ x d x = x ( l n x ) − x \int ln x dx = (ln x) x - \int 1/x * x dx = x(lnx)-x ∫ l n x d x = ( l n x ) x − ∫ 1 / x ∗ x d x = x ( l n x ) − x
35. ∫ 1 e x + e − x d x \int \frac{1}{e^x+e^{-x}} dx ∫ e x + e − x 1 d x
∫ 1 e x + e − x d x we know c o s h x = e x + e − x 2 = 1 2 ∫ 1 c o s h x d x = ∫ e x e 2 x + 1 d x ( u = e x , d u = e x d x ) = ∫ d u u 2 + 1 = arctan u = arctan e x + C
\begin{aligned}
&\int \frac{1}{e^x+e^{-x}} dx \\
& \text{we know} \; cosh x=\frac{e^x+e^{-x}}{2}\\
&=\frac{1}{2}\int \frac{1}{cosh x} dx \\
&=\int \frac{e^x}{e^{2x}+1} dx (u=e^x, du=e^xdx)\\
&=\int \frac{du}{u^2+1} =\arctan {u}\\
&=\arctan{e^x}+C
\end{aligned}
∫ e x + e − x 1 d x we know c o s h x = 2 e x + e − x = 2 1 ∫ c o s h x 1 d x = ∫ e 2 x + 1 e x d x ( u = e x , d u = e x d x ) = ∫ u 2 + 1 d u = arctan u = arctan e x + C
36. ∫ l o g 2 x d x \int log_2 x dx ∫ l o g 2 x d x
∫ l o g 2 x d x = ∫ l n x l n 2 d x = 1 l n 2 ∫ l n x d x = 1 l n 2 ( x l n x − x ) + C = x l o g 2 x − x l n 2 + C
\begin{aligned}
&\int log_2 x dx =\int \frac{ln x}{ln 2} dx\\
&=\frac{1}{ln 2}\int ln x dx\\
&=\frac{1}{ln 2}(x ln x - x)+C\\
&=x log_2x - \frac{x}{ln 2} + C\\
\end{aligned}
∫ l o g 2 x d x = ∫ l n 2 l n x d x = l n 2 1 ∫ l n x d x = l n 2 1 ( x l n x − x ) + C = x l o g 2 x − l n 2 x + C
37. ∫ x 3 ∗ s i n ( 2 x ) d x \int x^3*sin(2x) dx ∫ x 3 ∗ s i n ( 2 x ) d x
∫ x 3 sin 2 x d x = x 3 ( − 1 2 c o s 2 x ) − 3 x 2 ( − 1 4 s i n 2 x ) + 6 x ( 1 8 c o s 2 x ) − 6 ( 1 16 s i n 2 x ) = c o s 2 x ( − 1 2 x 3 + 3 4 x ) + s i n 2 x ( 3 4 x 2 − 3 8 ) + C
\begin{aligned}
&\int x^3\sin{2x} dx \\
&=x^3(-\frac{1}{2}cos2x)-3x^2(-\frac{1}{4}sin2x)+6x(\frac{1}{8}cos2x)-6(\frac{1}{16}sin2x)\\
&=cos 2x (-\frac{1}{2}x^3+\frac{3}{4}x)+sin2x(\frac{3}{4}x^2-\frac{3}{8})+C
\end{aligned}
∫ x 3 sin 2 x d x = x 3 ( − 2 1 c o s 2 x ) − 3 x 2 ( − 4 1 s i n 2 x ) + 6 x ( 8 1 c o s 2 x ) − 6 ( 1 6 1 s i n 2 x ) = c o s 2 x ( − 2 1 x 3 + 4 3 x ) + s i n 2 x ( 4 3 x 2 − 8 3 ) + C
38. ∫ x 2 [ 1 + x 3 ] 1 / 3 d x \int x^2[1+x^3]^{1/3} dx ∫ x 2 [ 1 + x 3 ] 1 / 3 d x
∫ x 2 1 + x 3 3 d x ( u = 1 + x 3 , d u = 3 x 2 d x ) = 1 3 ∫ u 3 d u = 1 3 ∫ u 1 3 d u = 1 3 3 4 u 1 + 1 3 = 1 4 u u 3 = 1 4 ( 1 + x 3 ) 1 + x 3 3 + C
\begin{aligned}
&\int x^2 \sqrt[3]{1+x^3} dx \\
&(u=1+x^3, du=3x^2dx) \\
&=\frac{1}{3}\int \sqrt[3]u du = \frac{1}{3}\int u^{\frac{1}{3}} du \\
&=\frac{1}{3} \frac{3}{4}u^{1+\frac{1}{3}}=\frac{1}{4}u\sqrt[3]{u}\\
&=\frac{1}{4}(1+x^3)\sqrt[3]{1+x^3}+C
\end{aligned}
∫ x 2 3 1 + x 3 d x ( u = 1 + x 3 , d u = 3 x 2 d x ) = 3 1 ∫ 3 u d u = 3 1 ∫ u 3 1 d u = 3 1 4 3 u 1 + 3 1 = 4 1 u 3 u = 4 1 ( 1 + x 3 ) 3 1 + x 3 + C
39. ∫ 1 / ( x 2 + 4 ) 2 d x \int 1/(x^2 + 4)^2 dx ∫ 1 / ( x 2 + 4 ) 2 d x
∫ 1 ( x 2 + 4 ) 2 d x ( x = 2 t a n y , d x = 2 s e c 2 y d y , y = arctan x 2 ) = ∫ 2 s e c 2 y ( 4 ( t a n 2 y + 1 ) ) 2 d y = ∫ s e c 2 y 8 s e c 4 y d y = 1 8 ∫ c o s 2 y d y = 1 16 ∫ 1 + cos 2 y d y = y 16 + 1 32 sin 2 y = 1 16 a r c t a n x 2 + 1 16 s i n y c o s y ( r i g h t t r i a n g l e a n g l e = y , h = s q r t ( x 2 + 4 ) a = 2 , o = x ) = 1 16 a r c t a n x 2 + 1 16 x x 2 + 4 2 x 2 + 4 = 1 16 a r c t a n x 2 + x 8 ( x 2 + 4 ) + C
\begin{aligned}
&\int \frac{1}{(x^2 + 4)^2} dx \\
&(x=2tany, dx=2sec^2ydy, y=\arctan{\frac{x}{2}}) \\
&=\int \frac{2sec^2y}{(4(tan^2y+1))^2}dy=\int \frac{sec^2y}{8sec^4y}dy\\
&=\frac{1}{8}\int cos^2y dy =\frac{1}{16}\int 1+\cos{2y}dy\\
&=\frac{y}{16}+\frac{1}{32}\sin{2y}=\frac{1}{16}arctan{\frac{x}{2}}+\frac{1}{16}sin y cos y\\
&(right triangle angle=y, h=sqrt(x^2+4) a=2, o=x)\\
&=\frac{1}{16}arctan{\frac{x}{2}}+\frac{1}{16}\frac{x}{\sqrt{x^2+4}} \frac{2}{\sqrt{x^2+4}}\\
&=\frac{1}{16}arctan{\frac{x}{2}}+\frac{x}{8(x^2+4)}+C
\end{aligned}
∫ ( x 2 + 4 ) 2 1 d x ( x = 2 t a n y , d x = 2 s e c 2 y d y , y = arctan 2 x ) = ∫ ( 4 ( t a n 2 y + 1 ) ) 2 2 s e c 2 y d y = ∫ 8 s e c 4 y s e c 2 y d y = 8 1 ∫ c o s 2 y d y = 1 6 1 ∫ 1 + cos 2 y d y = 1 6 y + 3 2 1 sin 2 y = 1 6 1 a r c t a n 2 x + 1 6 1 s i n y c o s y ( r i g h t t r i a n g l e a n g l e = y , h = s q r t ( x 2 + 4 ) a = 2 , o = x ) = 1 6 1 a r c t a n 2 x + 1 6 1 x 2 + 4 x x 2 + 4 2 = 1 6 1 a r c t a n 2 x + 8 ( x 2 + 4 ) x + C
40. ∫ 1 2 s q r t ( x 2 − 1 ) d x \int_1^2 sqrt(x^2-1) dx ∫ 1 2 s q r t ( x 2 − 1 ) d x
∫ 1 2 x 2 − 1 d x , ( x = s e c ( y ) , d x = s e c ( y ) t a n ( y ) d y ) ∫ tan 2 y sec y tan y d y = ∫ sec y tan 2 y d y ( s e c y t a n y − I − > s e c y ) = tan y sec y − ∫ sec 3 y d y
\begin{aligned}
&\int_1^2 \sqrt{x^2-1} dx , (x=sec(y), dx=sec(y) tan(y) dy)\\
&\int \sqrt{\tan^2y} \sec y \tan y dy\\
&=\int \sec y \tan^2 y dy (sec y tan y -I-> sec y) \\
&=\tan y \sec y -\int \sec^3 y dy\\
\end{aligned}
∫ 1 2 x 2 − 1 d x , ( x = s e c ( y ) , d x = s e c ( y ) t a n ( y ) d y ) ∫ tan 2 y sec y tan y d y = ∫ sec y tan 2 y d y ( s e c y t a n y − I − > s e c y ) = tan y sec y − ∫ sec 3 y d y
∫ sec 3 x d x = ∫ sec x sec 2 x d x ( s e c 2 x − I − > t a n x ) = sec x tan x − ∫ sec x tan x tan x d x = sec x tan x − ∫ sec x ( sec 2 x − 1 ) d x = sec x tan x − ∫ sec 3 x d x + ∫ sec x d x = sec x tan x + l n ∣ s e c x + t a n x ∣ − ∫ sec 3 x d x = 1 2 ( sec x tan x + l n ∣ s e c x + t a n x ∣ ) \int \sec^3 x dx = \int \sec x \sec^2 x dx (sec^2x-I->tan x)\\
=\sec x \tan x - \int \sec x \tan x \tan x dx\\
=\sec x \tan x - \int \sec x (\sec^2 x -1 ) dx \\
=\sec x \tan x - \int \sec^3 x dx +\int \sec x dx \\
=\sec x \tan x + ln |sec x + tan x|-\int \sec^3x dx\\
= \frac{1}{2}(\sec x \tan x + ln |sec x + tan x|)
∫ sec 3 x d x = ∫ sec x sec 2 x d x ( s e c 2 x − I − > t a n x ) = sec x tan x − ∫ sec x tan x tan x d x = sec x tan x − ∫ sec x ( sec 2 x − 1 ) d x = sec x tan x − ∫ sec 3 x d x + ∫ sec x d x = sec x tan x + l n ∣ s e c x + t a n x ∣ − ∫ sec 3 x d x = 2 1 ( sec x tan x + l n ∣ s e c x + t a n x ∣ )
x=sec(y), y=arcsec x, RT. angle=y, h=x, a=1, o=sqrt(x^2-1)= tan y sec y − ∫ sec 3 y d y = tan y sec y − 1 2 ( sec y tan y + l n ∣ s e c y + t a n y ∣ ) = 1 2 x x 2 − 1 − 1 2 l n ∣ x 2 − 1 + x ∣ + C ∫ 1 2 x 2 − 1 d x = [ 1 2 x x 2 − 1 − 1 2 l n ∣ x 2 − 1 + x ∣ ] 1 2 = 3 − 1 2 l n ( 3 + 2 )
=\tan y \sec y -\int \sec^3 y dy\\
=\tan y \sec y -\frac{1}{2}(\sec y \tan y + ln |sec y + tan y|)\\
=\frac{1}{2}x\sqrt{x^2-1}-\frac{1}{2}ln| \sqrt{x^2-1}+x|+C\\
\int_1^2 \sqrt{x^2-1} dx=\left[\frac{1}{2}x\sqrt{x^2-1}-\frac{1}{2}ln| \sqrt{x^2-1}+x|\right]_1^2\\
=\sqrt{3}-\frac{1}{2}ln(\sqrt{3}+2) \\
= tan y sec y − ∫ sec 3 y d y = tan y sec y − 2 1 ( sec y tan y + l n ∣ s e c y + t a n y ∣ ) = 2 1 x x 2 − 1 − 2 1 l n ∣ x 2 − 1 + x ∣ + C ∫ 1 2 x 2 − 1 d x = [ 2 1 x x 2 − 1 − 2 1 l n ∣ x 2 − 1 + x ∣ ] 1 2 = 3 − 2 1 l n ( 3 + 2 )
Author: crazyj7@gmail.com
31. [1:49:32](https://www.youtube.com/watch?v=dgm4-3-Iv3s&t=6572s) integral of (x-x^(3/2))^-1/2
32. [1:52:37](https://www.youtube.com/watch?v=dgm4-3-Iv3s&t=6757s) integral of (x-x^2)^-1/2
33. [1:56:03](https://www.youtube.com/watch?v=dgm4-3-Iv3s&t=6963s) integral of e^(2lnx)
34. [1:56:57](https://www.youtube.com/watch?v=dgm4-3-Iv3s&t=7017s) integral of lnx/sqrt x
35. [2:00:32](https://www.youtube.com/watch?v=dgm4-3-Iv3s&t=7232s) integral of 1/e^x+e^-x
36. [2:01:57](https://www.youtube.com/watch?v=dgm4-3-Iv3s&t=7317s) integral of log(x) base 2
37. [2:05:15](https://www.youtube.com/watch?v=dgm4-3-Iv3s&t=7515s) integral of x^3*sin2x
38. [2:08:32](https://www.youtube.com/watch?v=dgm4-3-Iv3s&t=7712s) integral of x^2[1+x^3]^1/3
39. [2:12:30](https://www.youtube.com/watch?v=dgm4-3-Iv3s&t=7950s) integral of 1/(x^2 + 4)^2
40. [2:19:38](https://www.youtube.com/watch?v=dgm4-3-Iv3s&t=8378s) integral of sqrt(x^2-1) from 1 to 2
