高等学校数学III 積分法/演習問題のソースを表示

== 問題 ==
次の不定積分を計算せよ。
#<math>\int \left(\frac{x+2}{x}\right)^2 dx</math>
#<math>\int \left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)dx</math>
#<math>\int \left(\sqrt{2x}+3\right)^2 dx</math>
#<math>\int \frac{x-\cos^2x}{x\cos^2x} dx</math>
#<math>\int \left(\tan^2x+\frac{1}{\tan^2x}\right) dx</math>
#<math>\int \left(e^x+\frac{1}{x}\right) dx</math>
#<math>\int 3^{x+2} dx</math>
#<math>\int (x+2)(x^2+4x+1)^3 dx</math>
#<math>\int \frac{x+1}{(3x-1)^3} dx</math>
#<math>\int \cos^3x \cdot \sin^2 x dx</math>
#<math>\int \frac{dx}{(1+\tan x)\cos^2x}</math>
#<math>\int (2x+1)e^{x^2+x+5} dx</math>
#<math>\int \frac{e^{4x}}{e^{2x}+1} dx</math>
#<math>\int x\sqrt{1-x} dx</math>
#<math>\int \frac{3x-1}{\sqrt{x+1}} dx</math>
#<math>\int \frac{x}{(x^2+4)^2} dx</math>
#<math>\int \frac{x^2+1}{x^4-5x^2+4} dx</math>
#<math>\int \frac{3x+2}{x(x+1)^3} dx</math>
#<math>\int \frac{x}{\sqrt[3]{x+1}-1} dx</math>
#<math>\int \frac{\cos x}{\sin x(\sin x+1)} dx</math>
#<math>\int \left(\tan x+\frac{1}{\tan x}\right)^2 dx</math>
#<math>\int \tan^4x dx</math>
#<math>\int \frac{\sin 2x}{1+\sin x} dx</math>
#<math>\int \frac{x}{1-\cos x} dx</math>
#<math>\int \frac{1}{1-\sin x} dx</math>
#<math>\int \frac{\sqrt{x}}{\sqrt[4]{x^3}+1} dx</math>
#<math>\int \log(x^2-1) dx</math>
#<math>\int \frac{e^{3x}}{e^x-1} dx</math>
#<math>\int \frac{e^x}{e^x-e^{-x}} dx</math>
#<math>\int \frac{dx}{3x^2-4x-2}</math>
#<math>\int \frac{dx}{\sqrt{2x^2-4x+3}}</math>
#<math>\int \frac{\tan x}{\cos^3 x}dx</math>
#<math>\int \frac{dx}{1+\cos x}</math>
#<math>\int x\tan^2x dx</math>
#<math>\int e^x \cos x dx</math>
#<math>\int \sqrt{x^2-x-1} dx</math>
#<math>\int \sin 2x \cdot \cos 3x dx</math>
#<math>\int \left(\sin x+\frac{1}{\sin x}\right)^2 dx</math>
#<math>\int \frac{1+\cos^3x}{\cos^2x}dx</math>
#<math>\int \frac{dx}{1+\sin x}</math>
#<math>\int \frac{dx}{e^x+1}</math>
#<math>\int \left(\log x\right)^2 dx</math>
#<math>\int \frac{\sqrt{1-x^2}}{x}dx</math>
#<math>\int \frac{dx}{x\sqrt{2x-x^2}}</math>

== 解答 ==
以下の解答においては、問iにおいて計算すべき積分をI<sub>i</sub>とおくことにする。

<math> 
\begin{align}
I_1&= \int \left(\frac{x+2}{x}\right)^2 dx \\
   &= \int \left(1+\frac{4}{x}+\frac{4}{x^2}\right) dx \\
   &= x+4\log|x|-\frac{4}{x}+C
\end{align}
</math>

<math> 
\begin{align}
I_2&= \int \left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right) dx \\
   &= \int \left(x-\sqrt{x}-2\right) dx \\ 
   &= \frac{x^2}{2}-\frac{2}{3}x\sqrt{x}-2x+C \\
\end{align}
</math>

<math> 
\begin{align}
I_3&=\int \left(\sqrt{2x}+3\right)^2 dx \\
   &=\int \left(2x+6\sqrt{2x}+9\right) dx \\
   &=x^2+4x\sqrt{2x}+9x+C
\end{align}
</math>

<math>
\begin{align}
I_4&= \int \frac{x-\cos^2x}{x\cos^2x} dx \\
   &= \int \left(\frac{1}{\cos^2x}-\frac{1}{x}\right) dx \\
   &= \tan x-\log|x|+C
\end{align}
</math>

<math>
\begin{align}
I_5&=\int \left(\tan^2x+\frac{1}{\tan^2x}\right) dx \\
   &=\int \left( \left(\frac{1}{\cos^2 x}-1\right)+\left(\frac{1}{\sin^2x}-1 \right)\right) dx\\
   &=\tan x-\frac{1}{\tan x}-2x+C
\end{align}
</math>

<math>
\begin{align}
I_6&=\int \left(e^x+\frac{1}{x}\right) dx \\
   &=e^x+\log|x|+C
\end{align}
</math>

<math>
\begin{align}
I_7&=\int 3^{x+2} dx \\
   &=\frac{3^{x+2}}{\log 3}+C
\end{align}
</math>

<math> 
\begin{align}
I_8&=\int (x+2)(x^2+4x+1)^3 dx \\
\end{align}
</math>

<math>t=x^2+4x+1</math>とおくと<math>dt=(2x+4)dx=2(x+2)dx</math>なので、

<math>
\begin{align}
I_8&=\int \frac{t^3}{2} dt \\
   &=\frac{t^4}{8}+C \\
   &=\frac{1}{8}(x^2+4x+1)^4+C
\end{align}
</math>

<math> 
\begin{align}
I_9&=\int \frac{x+1}{(3x-1)^3} dx\\
   &=\int \left(\frac{1}{3(3x-1)^2}+\frac{4}{3(3x-1)^3}\right) dx \\
   &=-\frac{1}{9(3x-1)}-\frac{2}{9(3x-1)^2}+C \\
   &=-\frac{3x+1}{9(3x-1)^2}+C
\end{align}
</math>

別解

<math>t=3x-1</math>と置く。
<math>x=\frac{t+1}{3},\;\;x+1=\frac{t+4}{3},\;\;\frac{dt}{dx}=3,\;\;dt=3dx</math>なので、

<math>
\begin{align}
I_{9}&=\frac{1}{9}\int \left(\frac{t+4}{t^3} \right)dt\\
     &=\frac{1}{9}\int \left(\frac{1}{t^2}+\frac{4}{t^3} \right)dt\\
     &=-\frac{1}{9} \left(\frac{1}{t}+\frac{2}{t^2} \right)+C\\
     &=-\frac{1}{9} \left(\frac{1}{3x-1}+\frac{2}{(3x-1)^2} \right)+C\\
     &=-\frac{3x+1}{9(3x-1)^2} +C
\end{align}
</math>

<math>
\begin{align}
I_{10}&=\int \cos^3x \cdot \sin^2 x dx \\
      &=\int \cos x(1-\sin^2x)\sin^2x dx \\
      &=\int \cos x(\sin^2x-\sin^4x) dx \\      
\end{align}</math>

<math>t=\sin x</math>とおくと<math>dt=\cos xdx</math>なので、

<math>
\begin{align}
I_{10}&=\int(t^2-t^4)dt \\
      &=\frac{t^3}{3}-\frac{t^5}{5}+C \\
      &=\frac{1}{3}\sin^3x-\frac{1}{5}\sin^5x+C
\end{align}
</math>

<math>
\begin{align}
I_{11}&=\int \frac{dx}{(1+\tan x)\cos^2x} \\
\end{align}</math>

<math>t=\tan x</math>とおくと<math>dt=\frac{dx}{\cos^2x}</math>なので、

<math>
\begin{align}
I_{11}&=\int \frac{dt}{1+t} \\
      &=\log |1+t|+C \\
      &=\log |1+\tan x|+C
\end{align}
</math>

<math>
\begin{align}
I_{12}&=\int (2x+1)e^{x^2+x+5} dx \\
\end{align}</math>

<math>t=x^2+x+5</math>とおくと<math>dt=(2x+1)dx</math>なので、

<math>
\begin{align}
I_{12}&=\int e^t dt \\
      &=e^t+C \\
      &=e^{x^2+x+5}+C
\end{align}
</math>

<math>
\begin{align}
I_{13}&=\int \frac{e^{4x}}{e^{2x}+1} dx \\
\end{align}</math>

<math>t=e^{2x}+1</math>とおくと<math>dt=2e^{2x}dx</math>なので、

<math>
\begin{align}
I_{13}&=\int \frac{t-1}{2t} dt \\
      &=\frac{1}{2} \int \left(1-\frac{1}{t}\right) dt \\
      &=\frac{1}{2}(t-\log t)+C' \\
      &=\frac{1}{2}\left(e^{2x}+1-\log(e^{2x}+1)\right)+C' \\
      &=\frac{1}{2}\left(e^{2x}-\log(e^{2x}+1)\right)+C
\end{align}
</math>

<math> 
\begin{align}
I_{14}&=\int x\sqrt{1-x} dx \\
\end{align}</math>

<math>t=\sqrt{1-x}</math>とおくと、<math>t^2=1-x,\;\;x=1-t^2,\;\;dx=-2tdt</math>なので、

<math>
\begin{align}
I_{14}&=-2\int (1-t^2)\cdot t\cdot t dt \\
      &=-2\int (t^2-t^4) dt \\
      &=-2\left(\frac{t^3}{3}-\frac{t^5}{5}\right)+C \\
      &=-\frac{2t^3}{15}(5-3t^2)+C \\
      &=-\frac{2}{15}(1-x)\sqrt{1-x}(5-3(1-x))+C \\
      &=-\frac{2}{15}(3x+2)(1-x)\sqrt{1-x}+C \\
\end{align}
</math>

<math>
\begin{align}
I_{15}&=\int \frac{3x-1}{\sqrt{x+1}} dx \\
\end{align}</math>

<math>t=\sqrt{x+1}</math>とおくと、<math>t^2=x+1,\;\;x=t^2-1,\;\;dx=2tdt</math>なので、

<math>
\begin{align}
I_{15}&=\int\frac{3(t^2-1)-1}{t}\cdot 2t dt \\ 
      &=2\int (3t^2-4) dt \\
      &=2(t^3-4t)+C \\
      &=2t(t^2-4)+C \\
      &=2((x+1)-4)\sqrt{x+1}+C \\
      &=2(x-3)\sqrt{x+1}+C \\
\end{align}
</math>

<math>
\begin{align}
I_{16}&=\int \frac{x}{(x^2+4)^2} dx \\
\end{align}</math>

<math>t=x^2+4</math>とおくと<math>dt=2xdx</math>なので、

<math>
\begin{align}
I_{16}&=\int \frac{dt}{2t^2} \\
      &=-\frac{1}{2t}+C \\
      &=-\frac{1}{2(x^2+4)}+C
\end{align}
</math>

<math> 
\begin{align}
I_{17}&=\int \frac{x^2+1}{x^4-5x^2+4} dx \\
      &=\frac{1}{3}\int \left(\frac{5}{x^2-4}-\frac{2}{x^2-1} \right) dx \\
      &=\frac{1}{3}\int \left(\frac{5}{4}\left(\frac{1}{x-2}-\frac{1}{x+2}\right)-\left(\frac{1}{x-1}-\frac{1}{x+1}\right)\right) dx \\
      &=\frac{5}{12}\log\left|\frac{x-2}{x+2}\right|-\frac{1}{3}\log\left|\frac{x-1}{x+1}\right|+C \\
      &=\frac{1}{3}\log\left|\frac{x+1}{x-1}\right|-\frac{5}{12}\log\left|\frac{x+2}{x-2}\right|+C \\
\end{align}
</math>

<math>
\begin{align}
I_{18}&=\int \frac{3x+2}{x(x+1)^3} dx \\
      &=\int \left(\frac{1}{(x+1)^3}+\frac{2}{x(x+1)^2}\right) dx \\
      &=\int \left(\frac{1}{(x+1)^3}-\frac{2}{(x+1)^2}+\frac{2}{x(x+1)}\right) dx \\
      &=\int \left(\frac{1}{(x+1)^3}-\frac{2}{(x+1)^2}+\frac{2}{x}-\frac{2}{x+1} \right) dx \\      
      &=-\frac{1}{2(x+1)^2}+\frac{2}{x+1}+2\log\left|\frac{x}{x+1}\right|+C \\
      &=2\log\left|\frac{x}{x+1}\right|+\frac{2}{x+1}-\frac{1}{2(x+1)^2}+C
\end{align}
</math>

<math>
\begin{align}
I_{19}&=\int \frac{x}{\sqrt[3]{x+1}-1} dx \\
      &=\int \frac{x\left(\left(\sqrt[3]{x+1}\right)^2+\sqrt[3]{x+1}+1\right)}{x} dx \\
      &=\int \left((x+1)^\frac{2}{3}+(x+1)^\frac{1}{3}+1\right) dx \\
      &=\frac{3}{5}(x+1)^\frac{5}{3}+\frac{3}{4}(x+1)^\frac{4}{3}+x+C \\
\end{align}
</math>

<math>
\begin{align}
I_{20}&=\int \frac{\cos x}{\sin x(\sin x+1)} dx \\
\end{align}</math>

<math>t=\sin x</math>とおくと<math>dt=\cos xdx</math>なので、

<math>
\begin{align}
I_{20}&=\int \frac{dt}{t(t+1)} \\
      &=\int \left(\frac{1}{t}-\frac{1}{t+1}\right) dt \\
      &=\log\left|\frac{t}{t+1}\right|+C \\
      &=\log\left|\frac{\sin x}{\sin x+1}\right|+C \\
\end{align}
</math>

<math>
\begin{align}
I_{21}&=\int \left(\tan x+\frac{1}{\tan x}\right)^2 dx \\
      &=\int\left(\tan^2x+2+\frac{1}{\tan^2x}\right) dx \\
      &=\int \left( \left(\frac{1}{\cos^2 x}-1\right)+2+\left(\frac{1}{\sin^2x}-1 \right)\right) dx\\
      &=\tan x-\frac{1}{\tan x}+C \\
\end{align}
</math>

<math>
\begin{align}
I_{22}&=\int \tan^4x dx \\
      &=\int \tan^2x \tan^2x dx \\
      &=\int \tan^2x \left(\frac{1}{\cos^2x}-1\right) dx \\
      &=\int \left(\frac{\tan^2x}{\cos^2x}-\tan^2x\right) dx \\
      &=\int \left(\frac{\tan^2x}{\cos^2x} -\left(\frac{1}{\cos^2x}-1\right)\right)dx  \\
      &=\int \frac{\tan^2x}{\cos^2x}dx - (\tan x-x) \\
\end{align}</math>

<math>t=\tan x</math>とおくと<math> dt=\frac{1}{\cos^2x}dx</math>なので、

<math>
\begin{align}
I_{22}&=\int t^2 dt - (\tan x-x) \\
      &=\frac{1}{3}t^3-\tan x+x+C \\
      &=\frac{1}{3}\tan^3x-\tan x+x+C
\end{align}
</math>

<math>
\begin{align}
I_{23}&=\int \frac{\sin 2x}{1+\sin x} dx \\
    &=\int \frac{2\sin x \cos x}{1+\sin x} dx 
\end{align}</math>

<math>t=\sin x</math>とおくと<math>dt=\cos xdx</math>なので、

<math>
\begin{align}
I_{23}&=\int \frac{2t}{1+t} dt \\
    &=2\int \left(1-\frac{1}{1+t}\right) dt \\
    &=2t-2\log|1+t|+C \\
    &=2\sin x-2\log\left(1+\sin x\right)+C
\end{align}
</math>

<math>
\begin{align}
I_{24}&=\int \frac{x}{1-\cos x} dx \\
      &=\int \frac{x}{2\sin^2 \frac{x}{2}} dx \\
      &=-\frac{x}{\tan \frac{x}{2}}+\int \frac{1}{\tan \frac{x}{2}} dx \\
      &=-\frac{x}{\tan \frac{x}{2}}+\int \frac{\cos \frac{x}{2}}{\sin \frac{x}{2}} dx 
\end{align}</math>

<math>t=\sin\frac{x}{2}</math>とおくと<math>dt=\frac{1}{2}\cos\frac{x}{2}dx</math>なので、

<math>
\begin{align}
I_{24}&=-\frac{x}{\tan \frac{x}{2}}+2\int \frac{dt}{t} \\
      &=-\frac{x}{\tan \frac{x}{2}}+2\log|t|+C \\
      &=-\frac{x}{\tan \frac{x}{2}}+2\log\left|\sin\frac{x}{2}\right|+C
\end{align}
</math>

<math>
\begin{align}
I_{25}&=\int \frac{1}{1-\sin x} dx \\
      &=\int \frac{1+\sin x}{1-\sin^2x}dx \\
      &=\int \left(\frac{1}{\cos^2 x}+\frac{\sin x}{\cos^2x}\right) dx \\
      &=\tan x+\int \frac{\sin x}{\cos^2x} dx
\end{align}</math>

<math>t=\cos x</math>とおくと<math>dt=-\sin xdx</math>なので、

<math>
\begin{align}
I_{25}&=\tan x-\int\frac{dt}{t^2} \\
      &=\tan x+\frac{1}{t}+C \\
      &=\tan x+\frac{1}{\cos x}+C      
\end{align}
</math>

<math>
\begin{align}
I_{26}&=\int \frac{\sqrt{x}}{\sqrt[4]{x^3}+1} dx
\end{align}</math>

<math>t=\sqrt[4]{x^3}</math>とおくと<math>dt=\frac{3}{4\sqrt[4]{x}}dx</math>なので、

<math>
\begin{align}
I_{26}&=\frac{4}{3}\int \frac{t}{t+1} dt \\
      &=\frac{4}{3}\int \left(1-\frac{1}{t+1}\right) dt \\
      &=\frac{4}{3}t-\frac{4}{3}\log|t+1|+C \\
      &=\frac{4}{3}\sqrt[4]{x^3}-\frac{4}{3}\log\left(\sqrt[4]{x^3}+1\right)+C
\end{align}
</math>

<math>
\begin{align}
I_{27}&=\int \log(x^2-1) dx \\
      &=\int \log(x+1)(x-1)  dx \\
      &=\int \left(\log|x+1|+\log|x-1|\right) dx \\
      &=(x+1)\log|x+1|-x+(x-1)\log|x-1|-x+C \\
      &=(x+1)\log|x+1|+(x-1)\log|x-1|-2x+C
\end{align}
</math>

<math>
\begin{align}
I_{28}&=\int \frac{e^{3x}}{e^x-1} dx
\end{align}</math>

<math>t=e^x</math>とおくと<math>dt=e^xdx</math>なので、

<math>
\begin{align}
I_{28}&=\int \frac{t^{2}}{t-1} dt \\
      &=\int \left(t+1+\frac{1}{t-1}\right) dt \\
      &=\frac{1}{2}t^2+t+\log|t-1|+C \\
      &=\frac{1}{2}e^{2x}+e^x+\log|e^x-1|+C
\end{align}
</math>

<math>
\begin{align}
I_{29}&=\int \frac{e^x}{e^x-e^{-x}} dx \\
\end{align}</math>

分子分母に<math>e^{x}</math>を掛けて、

<math>
\begin{align}
I_{29}&=\int \frac{e^{2x}}{e^{2x}-1} dx
\end{align}</math>

<math>t=e^{2x}</math>とおくと<math>dt=2e^{2x}dx</math>なので、

<math>
\begin{align}
I_{29}&=\frac{1}{2}\int \frac{dt}{t-1} \\
      &=\frac{1}{2}\log|t-1|+C \\
      &=\frac{1}{2}\log|e^{2x}-1|+C
\end{align}
</math>

<math>
\begin{align}
I_{30}&=\int \frac{dx}{3x^2-4x-2} \\
      &=\int \frac{3}{\left(3x-2-\sqrt{10}\right)\left(3x-2+\sqrt{10}\right)}dx \\
      &=\int \frac{3}{2\sqrt{10}}\left(\frac{1}{3x-2-\sqrt{10}}-\frac{1}{3x-2+\sqrt{10}}\right) dx \\
      &=\frac{1}{2\sqrt{10}} \left(\log|3x-2-\sqrt{10}|-\log|3x-2+\sqrt{10}|\right)+C \\
      &=\frac{1}{2\sqrt{10}} \log\left|\frac{3x-2-\sqrt{10}}{3x-2+\sqrt{10}}\right|+C
\end{align}
</math>

*<math>I_{30}</math>について補足
*:2行目から3行目への変形では、<math>\frac{1}{3x-2-\sqrt{10}}-\frac{1}{3x-2+\sqrt{10}}=\frac{(3x-2+\sqrt{10})-(3x-2-\sqrt{10})}{(3x-2-\sqrt{10})(3x-2+\sqrt{10})}=\frac{2\sqrt{10}}{(3x-2-\sqrt{10})(3x-2+\sqrt{10})}</math>であることに注意せよ。
*:3行目から4行目への変形では、<math>\left(\log|3x-2-\sqrt{10}|\right)'=\frac{3}{3x-2-\sqrt{10}},\ \left(\log|3x-2+\sqrt{10}|\right)'=\frac{3}{3x-2+\sqrt{10}}</math>であることに注意せよ。

<math>
\begin{align}
I_{31}&=\int \frac{dx}{\sqrt{2x^2-4x+3}} \\
      &=\frac{1}{\sqrt{2}}\int \frac{dx}{\sqrt{(x-1)^2+\frac{1}{2}}} \\
\end{align}</math>

<math>t=x-1+\sqrt{(x-1)^2+\frac{1}{2}}</math>とおくと<math>dt=\frac{x-1+\sqrt{(x-1)^2+\frac{1}{2}}}{\sqrt{(x-1)^2+\frac{1}{2}}}dx</math>なので、 

<math>\begin{align}
I_{31}&=\frac{1}{\sqrt{2}}\int \frac{dt}{t} \\
      &=\frac{1}{\sqrt{2}}\log|t|+C \\
      &=\frac{1}{\sqrt{2}}\log\left(x-1+\sqrt{(x-1)^2+\frac{1}{2}}\right)+C \\
      &=\frac{1}{\sqrt{2}}\log\left(x-1+\sqrt{x^2-2x+\frac{3}{2}}\right)+C
\end{align}
</math>

<math>
\begin{align}
I_{32}&=\int \frac{\tan x}{\cos^3 x}dx \\
      &=\int \frac{\sin x}{\cos^4 x}dx \\
\end{align}</math>

<math>t=\cos x</math>とおくと<math>dt=-\sin xdx</math>なので、

<math>
\begin{align}
I_{32}&=-\int \frac{dt}{t^4} \\
      &=\frac{1}{3t^3}+C \\
      &=\frac{1}{3\cos^3x}+C
\end{align}
</math>

<math>
\begin{align}
I_{33}&=\int \frac{dx}{1+\cos x} \\
      &=\int \frac{1-\cos x}{1-\cos^2x} dx \\
      &=\int \left(\frac{1}{\sin^2x}-\frac{\cos x}{\sin^2x} \right) dx \\
      &=-\frac{1}{\tan x}-\int \frac{\cos x}{\sin^2x} dx
\end{align}</math>

<math>t=\sin x</math>とおくと<math>dt=\cos xdx</math>なので、

<math>
\begin{align}
I_{33}&=-\frac{1}{\tan x}-\int \frac{dt}{t^2} \\
      &=-\frac{1}{\tan x}+\frac{1}{t}+C \\
      &=-\frac{1}{\tan x}+\frac{1}{\sin x}+C
\end{align}
</math>

別解

<math>
\begin{align}
I_{33}&=\int \frac{dx}{1+\cos x} \\
      &=\int \frac{dx}{2\cos^2 \frac{x}{2}} \\
      &=\tan\frac{x}{2}+C
\end{align}
</math>

<math>
\begin{align}
I_{34}&=\int x\tan^2x dx \\
      &=\int \left(\frac{x}{\cos^2x}-x\right) dx \\
      &=x\tan x-\int \tan x dx -\frac{1}{2}x^2\\
      &=x\tan x+\log|\cos x|-\frac{1}{2}x^2+C
\end{align}
</math>

<math>
\begin{align}
I_{35}&=\int e^x \cos x dx \\
      &=e^x \cos x+\int e^x \sin x dx \\
      &=e^x \cos x+e^x\sin x-\int e^x \cos x dx \\
      &=e^x \cos x+e^x\sin x-I_{35}
\end{align}</math>

なので、

<math>
\begin{align}
2I_{35}&=e^x \sin x+e^x\cos x+C' \\
 I_{35}&=\frac{e^x}{2}(\sin x+\cos x)+C
\end{align}
</math>

<math>
\begin{align}
I_{36}&=\int \sqrt{x^2-x-1} dx \\
      &=\int \sqrt{\left(x-\frac{1}{2}\right)^2-\frac{5}{4}}dx \\
\end{align}</math>

<math>t=x-\frac{1}{2}+\sqrt{\left(x-\frac{1}{2}\right)^2-\frac{5}{4}}</math>とすると<math>dt=\frac{x-\frac{1}{2}+\sqrt{\left(x-\frac{1}{2}\right)^2-\frac{5}{4}}}{\sqrt{\left(x-\frac{1}{2}\right)^2-\frac{5}{4}}} dx</math>なので、

<math>
\begin{align}
I_{36}&=\int \frac{\left(\frac{t}{2}+\frac{5}{8t}\right)^2-\frac{5}{4}}{t}dt \\
      &=\frac{1}{2}\int \left(\frac{t}{2}-\frac{5}{4t}+\frac{25}{32t^3}\right) dt \\
      &=\frac{1}{2}\left(\frac{t^2}{4}-\frac{5}{4}\log|t|-\frac{25}{64t^2}\right)+C \\
      &=\frac{1}{2}\left(\left(x-\frac{1}{2}\right)\sqrt{x^2-x-1}-\frac{5}{4}\log\left|x-\frac{1}{2}+\sqrt{x^2-x-1}\right|\right)+C
\end{align}
</math>

別解

<math>t=x-\frac{1}{2}+\sqrt{\left(x-\frac{1}{2}\right)^2-\frac{5}{4}}</math>と置く。…①

<math>\left \{ t-\left ( x-\frac{1}{2} \right ) \right \}^{2}=\left ( \sqrt{\left ( x-\frac{1}{2} \right )^{2}-\frac{5}{4}} \right )^{2}</math>

<math>2t\left ( x-\frac{1}{2} \right )=t^{2}+\frac{5}{4}</math>

<math>x-\frac{1}{2}=\frac{1}{2}\left ( t+\frac{5}{4t} \right )</math>…②

①②より、<math>\sqrt{\left ( x-\frac{1}{2} \right )^{2}-\frac{5}{4}}=t-\left ( x-\frac{1}{2} \right )
=t-\frac{1}{2}\left ( t+\frac{5}{4t} \right )
=\frac{1}{2}\left ( t-\frac{5}{4t} \right )</math>…③

②をtで微分すると、<math>\frac{dx}{dt}=\frac{1}{2}-\frac{5}{8t^{2}}=\frac{1}{2}\left ( 1-\frac{5}{4t^{2}} \right )</math>

<math>dx=\frac{1}{2}\left ( 1-\frac{5}{4t^{2}}\right )dt</math>…④

②③④より

<math>
\begin{align}
I_{36}&=\int \frac{1}{2} \left ( t-\frac{5}{4t} \right )\cdot \frac{1}{2}\left ( 1-\frac{5}{4t^{2}} \right )dt\\
&=\frac{1}{4}\int \left ( t-\frac{10}{4t} +\frac{25}{16t^{3}}\right )dt\\
&=\frac{1}{4}\left ( \frac{t^{2}}{2}-\frac{5}{2}\log\left | t \right | -\frac{25}{32t^{2}} \right )+C\\
&=\frac{1}{8}\left ( t^{2} -\frac{25}{16t^{2}}\right )-\frac{5}{8}\log \left | t \right |+C\\
&=\frac{1}{2}\cdot \frac{1}{2}\left ( t+\frac{5}{4t} \right )\frac{1}{2}\left ( t-\frac{5}{4t} \right )-\frac{5}{8}\log \left | t \right |+C\\
&=\frac{1}{2}\left ( x-\frac{1}{2} \right )\sqrt{x^{2}-x-1}-\frac{5}{8}\log \left | x-\frac{1}{2}+\sqrt{x^{2}-x-1} \right |+C\\
\end{align}
</math>

<math>
\begin{align}
I_{37}&=\int \sin 2x \cdot \cos 3x dx \\
      &=\frac{1}{2}\int (\sin 5x-\sin x)dx \\
      &=-\frac{1}{10}\cos 5x+\frac{1}{2}\cos x +C
\end{align}
</math>

<math>
\begin{align}
I_{38}&=\int \left(\sin x+\frac{1}{\sin x}\right)^2 dx \\
      &=\int \left(\sin^2x+2+\frac{1}{\sin^2x}\right) dx \\
      &=\int \left(\frac{1-\cos 2x}{2}+2+\frac{1}{\sin^2x}\right) dx \\
      &=\frac{5}{2}x-\frac{\sin 2x}{4}-\frac{1}{\tan x}+C
\end{align}
</math>

<math>
\begin{align}
I_{39}&=\int \frac{1+\cos^3x}{\cos^2x}dx \\
      &=\int \left(\frac{1}{\cos^2x}+\cos x\right) dx \\
      &=\tan x+\sin x+C
\end{align}
</math>

<math>
\begin{align}
I_{40}&=\int \frac{dx}{1+\sin x} \\
      &=\int \frac{1-\sin x}{1-\sin^2 x} dx \\
      &=\int \left(\frac{1}{\cos^2x}-\frac{\sin x}{\cos^2x}\right) dx \\
      &=\tan x-\int\frac{\sin x}{\cos^2x} dx \\
\end{align}</math>

<math>t=\cos x</math>とおくと<math>dt=-\sin xdx</math>なので、

<math>
\begin{align}
I_{40}&=\tan x+\int\frac{dt}{t^2} \\
      &=\tan x-\frac{1}{t}+C \\
      &=\tan x-\frac{1}{\cos x}+C
\end{align}
</math>

<math>
\begin{align}
I_{41}&=\int \frac{dx}{e^x+1}
\end{align}</math>

<math>t=e^x</math>とおくと<math>dt=e^xdx</math>なので、

<math>
\begin{align}
I_{41}&=\int \frac{dt}{t(t+1)} \\ 
      &=\int \left(\frac{1}{t}-\frac{1}{t+1}\right)dt \\
      &=\log|t|-\log|t+1|+C \\
      &=x-\log(e^x+1)+C
\end{align}
</math>

<math>
\begin{align}
I_{42}&=\int \left(\log x\right)^2 dx \\
      &=x\left(\log x\right)^2-2\int \log xdx \\
      &=x\left(\log x\right)^2-2x\log x+2x+C
\end{align}
</math>

<math>
\begin{align}
I_{43}&=\int \frac{\sqrt{1-x^2}}{x}dx
\end{align}</math>

<math>t=\sqrt{1-x^2}</math>とおくと<math>dt=-\frac{x}{\sqrt{1-x^2}}dx</math>なので、

<math>
\begin{align}
I_{43}&=\int \frac{t^2}{t^2-1}dt \\
      &=\int \left(1+\frac{1}{t^2-1}\right) dt \\
      &=\int \left(1+\frac{1}{2(t-1)}-\frac{1}{2(t+1)}\right) dt \\
      &=t+\log\sqrt{\left|\frac{t-1}{t+1}\right|}+C \\
      &=\sqrt{1-x^2}+\log\frac{1-\sqrt{1-x^2}}{|x|}+C
\end{align}
</math>

<math>
\begin{align}
I_{44}&=\int \frac{dx}{x\sqrt{2x-x^2}} \\
\end{align}</math>

<math>t=\frac{1}{x}</math>とおくと<math>dt=-\frac{1}{x^2}dx</math>なので、

<math>
\begin{align}
I_{44}&=-\int \frac{dt}{\sqrt{2t-1}} \\
      &=-\sqrt{2t-1}+C \\
      &=-\sqrt\frac{2-x}{x}+C
\end{align}</math>

[[カテゴリ:高等学校数学III]]