Integral100 [71-80]

71. $\int 1/(cbrt(x)+1)dx$

$\begin{aligned} &\int \frac{1}{\sqrt[3]{x}+1}dx \\ &(u=\sqrt[3]{x}, du=\frac{1}{3\sqrt[3]{x^2}}dx=\frac{1}{3}x^{-\frac{2}{3}}dx=\frac{1}{3}u^{-2}dx)\\ &=3\int \frac{u^2}{u+1}du=3\int \frac{u^2-1+1}{u+1}du\\ &=3\int \frac{(u-1)(u+1)}{u+1}du+3\int\frac{1}{u+1}du \\ &=3(\frac{1}{2}u^2-u)+3ln|u+1|\\ &=\frac{3}{2}u^2-3u+3ln|u+1|+C\\ &=\frac{3}{2}\sqrt[3]{x}^2-3\sqrt[3]{x}+3ln|\sqrt[3]{x}+1|+C\\ \end{aligned}$
Alt.
$(u=\sqrt[3]{x}+1, x=(u-1)^3, dx=3(u-1)^2du)\\ =\int \frac{1}{u} 3(u-1)^2du\\ =3\int \frac{u^2-2u+1}{u} du=3\int u-2+(1/u) du\\ =3(\frac{1}{2}u^2-2u+ln|u|)\\ =\frac{3}{2}(\sqrt[3]{x}+1)^2-6(\sqrt[3]{x}+1)+3ln|\sqrt[3]{x}+1|+C$

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72. $\int \frac{1}{cbrt(x + 1)}dx$

$\begin{aligned} &\int \frac{1}{\sqrt[3]{x+1}}dx \\ &(u=\sqrt[3]{x+1}, du=\frac{1}{3\sqrt[3]{(x+1)^2}}dx=\frac{1}{3}(x+1)^{-\frac{2}{3}}dx=\frac{1}{3}u^{-2}dx)\\ &=3\int \frac{u^2}{u}du=3\int u du \\ &=\frac{3}{2}u^2+C\\ &=\frac{3}{2}(x+1)^{\frac{2}{3}}+C \end{aligned}$
Alt.
(u=x+1)
$\begin{aligned} &\int \frac{1}{\sqrt[3]{x+1}}dx \\ &=\int u^{-\frac{1}{3}}du \\ &=\frac{3}{2}u^2+C\\ &=\frac{3}{2}(x+1)^{\frac{2}{3}}+C \end{aligned}$

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73. $\int (sin(x)+cos(x))^2 dx$

$\begin{aligned} &\int (sin(x)+cos(x))^2dx \\ &=\int 1+2sin(x)cos(x)dx=x+\int sin(2x) dx\\ &=x-\frac{1}{2}cos(2x)+C \end{aligned}$

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74. $\int 2xln(1+x) dx$

$\begin{aligned} &\int 2xln(1+x) dx =2\int xln(1+x) dx\\ & \text{(ln쪽을 미분하는 방향으로 부분적분)}\\ &=2( ln(1+x)\frac{1}{2}x^2-\int \frac{1}{1+x}\frac{1}{2}x^2dx )\\ &= ln(1+x)x^2-\int \frac{x^2}{1+x}dx = ln(1+x)x^2-\int \frac{x^2-1+1}{1+x}dx\\ &= ln(1+x)x^2-\int x-1+\frac{1}{1+x}dx\\ &= ln(1+x)x^2-\frac{1}{2}x^2+x-ln(1+x)+C\\ \end{aligned}$

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75. $\int 1/(x(1+sin^2ln(x)))dx$

$\begin{aligned} &\int \frac{1}{x(1+sin^2ln(x))} dx \\ &( u=ln(x), du=\frac{1}{x}dx : 1/x과 ln(x)에서 치환 착안.)\\ &=\int \frac{du}{1+sin^2u} : (sin에 대해 치환, 변환(반각)해보고, cos으로 나눠도 본다.\\ & sec, tan을 만들 수 있다는 것을 착안.)\\ &=\int \frac{sec^2u}{sec^2u+tan^2u}du\\ &=\int \frac{sec^2u}{1+2tan^2u}du, (t=tanu, dt=sec^2udu)\\ &=\int \frac{dt}{1+2t^2}, (t=tanu, dt=sec^2udu)\\ &=\frac{1}{\sqrt{2}}arctan {\sqrt{2}t}=\frac{1}{\sqrt{2}}arctan({\sqrt{2} tan(ln(x))})+C \end{aligned}$

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76. $\int sqrt((1-x)/(1+x))dx$

Try
$\begin{aligned} &\int \sqrt{\frac{1-x}{1+x}} dx \\ &(\text{전체치환? 일부치환. 분자분모곱, pm 등})\\ &( u = \sqrt{\frac{1-x}{1+x}}, du=\frac{1}{2}\sqrt{\frac{1+x}{1-x}}\frac{1-x+1+x}{1-2x+x^2}dx)\\ &( du = \frac{1}{2}\sqrt{\frac{1+x}{1-x}}\frac{2}{(1-x)^2}dx=\frac{1}{u(1-x)^2}dx)\\ &(u^2=\frac{1-x}{1+x}, xu^2+x=1-u^2, x=\frac{1-u^2}{1+u^2} )\\ &=\int u^2(1-x)^2 du =\int u^2(1-\frac{1-u^2}{1+u^2})^2du\\ &=\int u^2(\frac{2u^2}{1+u^2})^2 du = \int 4u^6(\frac{1}{1+u^2})^2du\\ \end{aligned}$
Try
$\begin{aligned} &\int \sqrt{\frac{1-x}{1+x}} dx \\ &(u=\frac{1-x}{1+x}, du=\frac{-1-x-(1-x)}{1+2x+x^2}dx=\frac{-2}{(1+x)^2}dx)\\ &( xu+u=1-x, x(u+1)=1-u, x=\frac{1-u}{1+u})\\ &=-\frac{1}{2}\int \sqrt{u} (1+x^2) du = -\frac{1}{2}\int \sqrt{u} (1+(\frac{1-2u+u^2}{1+2u+u^2})^2) du \\ &=-\frac{1}{2}\int \sqrt{u} (\frac{2+2u^2}{1+2u+u^2})^2 du \\ &=-2\int\sqrt{u}\frac{(1+u^2)^2}{(1+u)^4}du \end{aligned}$
Solve
$\begin{aligned} &\int \sqrt{\frac{1-x}{1+x}} dx =\int \sqrt{\frac{(1-x)(1-x)}{(1+x)(1-x)}} dx\\ &=\int \sqrt{\frac{(1-x)^2}{1-x^2}} dx =\int \frac{1-x}{\sqrt{1-x^2}}dx\\ &=\int \frac{1}{\sqrt{1-x^2}}dx-\int \frac{x}{\sqrt{1-x^2}}dx\\ & (u=1-x^2, du=-2xdx)\\ &=\sin^{-1}x+\frac{1}{2}\int\frac{1}{\sqrt{u}}du=\sin^{-1}x+\frac{1}{2}\int u^{-\frac{1}{2}}du\\ &=\sin^{-1}x+\sqrt{1-x^2}+C \end{aligned}$

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77. $\int x^{x/ln(x)}dx$

Interesting…
$\begin{aligned} &\int x^{\frac{x}{ln(x)}} dx\\ &( (\frac{x}{ln(x)})'=\frac{ln(x)-1}{(ln(x))^2})\\ &(y=x^{\frac{x}{ln(x)}}, ln(y)=\frac{x}{ln(x)}ln(x))\\ &( \frac{1}{y}y'= \frac{ln(x)-1}{(ln(x))^2}ln(x)+\frac{x}{ln(x)}\frac{1}{x})\\ &( \frac{1}{y}y'= \frac{ln(x)-1}{ln(x)}+\frac{1}{ln(x)}=1)\\ &y'=y\\ &=x^{\frac{x}{ln(x)}}+C \end{aligned}$
Alt.
$\begin{aligned} &\int x^{\frac{x}{ln(x)}} dx = \int (e^{ln x})^{\frac{x}{ln(x)}}dx=\int e^x dx\\ &= e^x+C\\ &(x^{\frac{x}{ln(x)}}=e^x) \end{aligned}$

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78. $\int arcsin(sqrt(x))dx$

$\begin{aligned} &\int arcsin(\sqrt x) dx \\ &( \text{ 루트부분을 치환. 부분적분} )\\ &( u = \sqrt x, u^2=x, 2udu=dx)\\ &=2\int u sin^{-1}(u) du= (arcsin은 미분가능)\\ &=2( sin^{-1}u \frac{1}{2}u^2-\int \frac{1}{\sqrt{1-u^2}}\frac{1}{2}u^2 du )\\ &=u^2sin^{-1}u-\int \frac{u^2}{\sqrt{1-u^2}}du\\ &=u^2sin^{-1}u+\int \frac{-1+1-u^2}{\sqrt{1-u^2}}du\\ &=x\sin^{-1}\sqrt{x}-\int \frac{1}{\sqrt{1-u^2}}du+\int \sqrt{1-u^2}du\\ &=x\sin^{-1}\sqrt{x}-\sin^{-1}\sqrt{x}+\int \sqrt{1-u^2}du\\ \end{aligned}$

$\int \sqrt{1-u^2}du, (u=sin\theta, du=cos\theta d\theta)\\ =\int cos^2\theta d\theta=\frac{1}{2}\int1+cos2\theta d\theta=\frac{1}{2}\theta+\frac{1}{2}\int cos 2\theta d\theta\\ =\frac{1}{2}\theta+\frac{1}{4} sin 2\theta =\frac{1}{2}sin^{-1}u+\frac{1}{2} sin \theta cos \theta\\ =\frac{1}{2}sin^{-1}u+\frac{1}{2} u \sqrt{1-u^2}$

$\begin{aligned} &\int arcsin(\sqrt x) dx \\ &=x\sin^{-1}\sqrt{x}-\sin^{-1}\sqrt{x}+\int \sqrt{1-u^2}du\\ &=x\sin^{-1}\sqrt{x}-\sin^{-1}\sqrt{x}+\frac{1}{2}sin^{-1}u+\frac{1}{2} u \sqrt{1-u^2}\\ &=x\sin^{-1}\sqrt{x}-\sin^{-1}\sqrt{x}+\frac{1}{2}sin^{-1}\sqrt{x}+\frac{1}{2} \sqrt{x} \sqrt{1-x}\\ &=(\sin^{-1}x)(x-\frac{1}{2})+\frac{1}{2}\sqrt{x(1-x)}+C \end{aligned}$

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79. $\int arctan(x) dx$

$\begin{aligned} &\int arctan(x) dx \\ &(y=arctan(x), x=tan(y), dx=sec^2ydy)\\ &=\int y sec^2y dy (부분적분)\\ &=y \tan(y)-\int tan(y) dy=x\arctan{x}+ln|cos y|+C\\ &=x\arctan{x}+ln|\frac{1}{\sqrt{1+x^2}}|+C\\ \end{aligned}$
Alt.
$\begin{aligned} &\int arctan(x) dx \\ &(\text{we know} \int \frac{1}{1+x^2} dx=arctan(x))\\ &=arctan(x) x -\int \frac{x}{1+x^2} dx (x^2을 치환)\\ &=x \arctan{x}-\frac{1}{2}ln|1+x^2|+C \end{aligned}$

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80. $\int_0^5 f(x) dx$

if x<=2, f(x)=10
if x>=2, f(x)=$3x^2-2$
$\begin{aligned} &=\int_0^2 f(x)dx+ \int_2^5 f(x) dx \\ &=\int_0^2 10 dx+ \int_2^5 3x^2-2 dx \\ &=20+\bigg[x^3-2x \bigg ]_2^5=20+(115-4)\\ &=131 \end{aligned}$

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