高等学校数学III 積分法/演習問題

問題

次の不定積分を計算せよ。

$\int \left({\frac {x+2}{x}}\right)^{2}dx$
$\int \left({\sqrt {x}}+1\right)\left({\sqrt {x}}-2\right)dx$
$\int \left({\sqrt {2x}}+3\right)^{2}dx$
$\int {\frac {x-\cos ^{2}x}{x\cos ^{2}x}}dx$
$\int \left(\tan ^{2}x+{\frac {1}{\tan ^{2}x}}\right)dx$
$\int \left(e^{x}+{\frac {1}{x}}\right)dx$
$\int 3^{x+2}dx$
$\int (x+2)(x^{2}+4x+1)^{3}dx$
$\int {\frac {x+1}{(3x-1)^{3}}}dx$
$\int \cos ^{3}x\cdot \sin ^{2}xdx$
$\int {\frac {dx}{(1+\tan x)\cos ^{2}x}}$
$\int (2x+1)e^{x^{2}+x+5}dx$
$\int {\frac {e^{4x}}{e^{2x}+1}}dx$
$\int x{\sqrt {1-x}}dx$
$\int {\frac {3x-1}{\sqrt {x+1}}}dx$
$\int {\frac {x}{(x^{2}+4)^{2}}}dx$
$\int {\frac {x^{2}+1}{x^{4}-5x^{2}+4}}dx$
$\int {\frac {3x+2}{x(x+1)^{3}}}dx$
$\int {\frac {x}{{\sqrt[{3}]{x+1}}-1}}dx$
$\int {\frac {\cos x}{\sin x(\sin x+1)}}dx$
$\int \left(\tan x+{\frac {1}{\tan x}}\right)^{2}dx$
$\int \tan ^{4}xdx$
$\int {\frac {\sin 2x}{1+\sin x}}dx$
$\int {\frac {x}{1-\cos x}}dx$
$\int {\frac {1}{1-\sin x}}dx$
$\int {\frac {\sqrt {x}}{{\sqrt[{4}]{x^{3}}}+1}}dx$
$\int \log(x^{2}-1)dx$
$\int {\frac {e^{3x}}{e^{x}-1}}dx$
$\int {\frac {e^{x}}{e^{x}-e^{-x}}}dx$
$\int {\frac {dx}{3x^{2}-4x-2}}$
$\int {\frac {dx}{\sqrt {2x^{2}-4x+3}}}$
$\int {\frac {\tan x}{\cos ^{3}x}}dx$
$\int {\frac {dx}{1+\cos x}}$
$\int x\tan ^{2}xdx$
$\int e^{x}\cos xdx$
$\int {\sqrt {x^{2}-x-1}}dx$
$\int \sin 2x\cdot \cos 3xdx$
$\int \left(\sin x+{\frac {1}{\sin x}}\right)^{2}dx$
$\int {\frac {1+\cos ^{3}x}{\cos ^{2}x}}dx$
$\int {\frac {dx}{1+\sin x}}$
$\int {\frac {dx}{e^{x}+1}}$
$\int \left(\log x\right)^{2}dx$
$\int {\frac {\sqrt {1-x^{2}}}{x}}dx$
$\int {\frac {dx}{x{\sqrt {2x-x^{2}}}}}$

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_iとおくことにする。

${\begin{aligned}I_{1}&=\int \left({\frac {x+2}{x}}\right)^{2}dx\\&=\int \left(1+{\frac {4}{x}}+{\frac {4}{x^{2}}}\right)dx\\&=x+4\log |x|-{\frac {4}{x}}+C\end{aligned}}$

${\begin{aligned}I_{2}&=\int \left({\sqrt {x}}+1\right)\left({\sqrt {x}}-2\right)dx\\&=\int \left(x-{\sqrt {x}}-2\right)dx\\&={\frac {x^{2}}{2}}-{\frac {2}{3}}x{\sqrt {x}}-2x+C\\\end{aligned}}$

${\begin{aligned}I_{3}&=\int \left({\sqrt {2x}}+3\right)^{2}dx\\&=\int \left(2x+6{\sqrt {2x}}+9\right)dx\\&=x^{2}+4x{\sqrt {2x}}+9x+C\end{aligned}}$

${\begin{aligned}I_{4}&=\int {\frac {x-\cos ^{2}x}{x\cos ^{2}x}}dx\\&=\int \left({\frac {1}{\cos ^{2}x}}-{\frac {1}{x}}\right)dx\\&=\tan x-\log |x|+C\end{aligned}}$

${\begin{aligned}I_{5}&=\int \left(\tan ^{2}x+{\frac {1}{\tan ^{2}x}}\right)dx\\&=\int \left(\left({\frac {1}{\cos ^{2}x}}-1\right)+\left({\frac {1}{\sin ^{2}x}}-1\right)\right)dx\\&=\tan x-{\frac {1}{\tan x}}-2x+C\end{aligned}}$

${\begin{aligned}I_{6}&=\int \left(e^{x}+{\frac {1}{x}}\right)dx\\&=e^{x}+\log |x|+C\end{aligned}}$

${\begin{aligned}I_{7}&=\int 3^{x+2}dx\\&={\frac {3^{x+2}}{\log 3}}+C\end{aligned}}$

${\begin{aligned}I_{8}&=\int (x+2)(x^{2}+4x+1)^{3}dx\\\end{aligned}}$

$t=x^{2}+4x+1$ とおくと $dt=(2x+4)dx=2(x+2)dx$ なので、

${\begin{aligned}I_{8}&=\int {\frac {t^{3}}{2}}dt\\&={\frac {t^{4}}{8}}+C\\&={\frac {1}{8}}(x^{2}+4x+1)^{4}+C\end{aligned}}$

${\begin{aligned}I_{9}&=\int {\frac {x+1}{(3x-1)^{3}}}dx\\&=\int \left({\frac {1}{3(3x-1)^{2}}}+{\frac {4}{3(3x-1)^{3}}}\right)dx\\&=-{\frac {1}{9(3x-1)}}-{\frac {2}{9(3x-1)^{2}}}+C\\&=-{\frac {3x+1}{9(3x-1)^{2}}}+C\end{aligned}}$

別解

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

${\begin{aligned}I_{9}&={\frac {1}{9}}\int \left({\frac {t+4}{t^{3}}}\right)dt\\&={\frac {1}{9}}\int \left({\frac {1}{t^{2}}}+{\frac {4}{t^{3}}}\right)dt\\&=-{\frac {1}{9}}\left({\frac {1}{t}}+{\frac {2}{t^{2}}}\right)+C\\&=-{\frac {1}{9}}\left({\frac {1}{3x-1}}+{\frac {2}{(3x-1)^{2}}}\right)+C\\&=-{\frac {3x+1}{9(3x-1)^{2}}}+C\end{aligned}}$

${\begin{aligned}I_{10}&=\int \cos ^{3}x\cdot \sin ^{2}xdx\\&=\int \cos x(1-\sin ^{2}x)\sin ^{2}xdx\\&=\int \cos x(\sin ^{2}x-\sin ^{4}x)dx\\\end{aligned}}$

$t=\sin x$ とおくと $dt=\cos xdx$ なので、

${\begin{aligned}I_{10}&=\int (t^{2}-t^{4})dt\\&={\frac {t^{3}}{3}}-{\frac {t^{5}}{5}}+C\\&={\frac {1}{3}}\sin ^{3}x-{\frac {1}{5}}\sin ^{5}x+C\end{aligned}}$

${\begin{aligned}I_{11}&=\int {\frac {dx}{(1+\tan x)\cos ^{2}x}}\\\end{aligned}}$

$t=\tan x$ とおくと $dt={\frac {dx}{\cos ^{2}x}}$ なので、

${\begin{aligned}I_{11}&=\int {\frac {dt}{1+t}}\\&=\log |1+t|+C\\&=\log |1+\tan x|+C\end{aligned}}$

${\begin{aligned}I_{12}&=\int (2x+1)e^{x^{2}+x+5}dx\\\end{aligned}}$

$t=x^{2}+x+5$ とおくと $dt=(2x+1)dx$ なので、

${\begin{aligned}I_{12}&=\int e^{t}dt\\&=e^{t}+C\\&=e^{x^{2}+x+5}+C\end{aligned}}$

${\begin{aligned}I_{13}&=\int {\frac {e^{4x}}{e^{2x}+1}}dx\\\end{aligned}}$

$t=e^{2x}+1$ とおくと $dt=2e^{2x}dx$ なので、

${\begin{aligned}I_{13}&=\int {\frac {t-1}{2t}}dt\\&={\frac {1}{2}}\int \left(1-{\frac {1}{t}}\right)dt\\&={\frac {1}{2}}(t-\log t)+C'\\&={\frac {1}{2}}\left(e^{2x}+1-\log(e^{2x}+1)\right)+C'\\&={\frac {1}{2}}\left(e^{2x}-\log(e^{2x}+1)\right)+C\end{aligned}}$

${\begin{aligned}I_{14}&=\int x{\sqrt {1-x}}dx\\\end{aligned}}$

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

${\begin{aligned}I_{14}&=-2\int (1-t^{2})\cdot t\cdot tdt\\&=-2\int (t^{2}-t^{4})dt\\&=-2\left({\frac {t^{3}}{3}}-{\frac {t^{5}}{5}}\right)+C\\&=-{\frac {2t^{3}}{15}}(5-3t^{2})+C\\&=-{\frac {2}{15}}(1-x){\sqrt {1-x}}(5-3(1-x))+C\\&=-{\frac {2}{15}}(3x+2)(1-x){\sqrt {1-x}}+C\\\end{aligned}}$

${\begin{aligned}I_{15}&=\int {\frac {3x-1}{\sqrt {x+1}}}dx\\\end{aligned}}$

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

${\begin{aligned}I_{15}&=\int {\frac {3(t^{2}-1)-1}{t}}\cdot 2tdt\\&=2\int (3t^{2}-4)dt\\&=2(t^{3}-4t)+C\\&=2t(t^{2}-4)+C\\&=2((x+1)-4){\sqrt {x+1}}+C\\&=2(x-3){\sqrt {x+1}}+C\\\end{aligned}}$

${\begin{aligned}I_{16}&=\int {\frac {x}{(x^{2}+4)^{2}}}dx\\\end{aligned}}$

$t=x^{2}+4$ とおくと $dt=2xdx$ なので、

${\begin{aligned}I_{16}&=\int {\frac {dt}{2t^{2}}}\\&=-{\frac {1}{2t}}+C\\&=-{\frac {1}{2(x^{2}+4)}}+C\end{aligned}}$

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

${\begin{aligned}I_{18}&=\int {\frac {3x+2}{x(x+1)^{3}}}dx\\&=\int \left({\frac {1}{(x+1)^{3}}}+{\frac {2}{x(x+1)^{2}}}\right)dx\\&=\int \left({\frac {1}{(x+1)^{3}}}-{\frac {2}{(x+1)^{2}}}+{\frac {2}{x(x+1)}}\right)dx\\&=\int \left({\frac {1}{(x+1)^{3}}}-{\frac {2}{(x+1)^{2}}}+{\frac {2}{x}}-{\frac {2}{x+1}}\right)dx\\&=-{\frac {1}{2(x+1)^{2}}}+{\frac {2}{x+1}}+2\log \left|{\frac {x}{x+1}}\right|+C\\&=2\log \left|{\frac {x}{x+1}}\right|+{\frac {2}{x+1}}-{\frac {1}{2(x+1)^{2}}}+C\end{aligned}}$

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

${\begin{aligned}I_{20}&=\int {\frac {\cos x}{\sin x(\sin x+1)}}dx\\\end{aligned}}$

$t=\sin x$ とおくと $dt=\cos xdx$ なので、

${\begin{aligned}I_{20}&=\int {\frac {dt}{t(t+1)}}\\&=\int \left({\frac {1}{t}}-{\frac {1}{t+1}}\right)dt\\&=\log \left|{\frac {t}{t+1}}\right|+C\\&=\log \left|{\frac {\sin x}{\sin x+1}}\right|+C\\\end{aligned}}$

${\begin{aligned}I_{21}&=\int \left(\tan x+{\frac {1}{\tan x}}\right)^{2}dx\\&=\int \left(\tan ^{2}x+2+{\frac {1}{\tan ^{2}x}}\right)dx\\&=\int \left(\left({\frac {1}{\cos ^{2}x}}-1\right)+2+\left({\frac {1}{\sin ^{2}x}}-1\right)\right)dx\\&=\tan x-{\frac {1}{\tan x}}+C\\\end{aligned}}$

${\begin{aligned}I_{22}&=\int \tan ^{4}xdx\\&=\int \tan ^{2}x\tan ^{2}xdx\\&=\int \tan ^{2}x\left({\frac {1}{\cos ^{2}x}}-1\right)dx\\&=\int \left({\frac {\tan ^{2}x}{\cos ^{2}x}}-\tan ^{2}x\right)dx\\&=\int \left({\frac {\tan ^{2}x}{\cos ^{2}x}}-\left({\frac {1}{\cos ^{2}x}}-1\right)\right)dx\\&=\int {\frac {\tan ^{2}x}{\cos ^{2}x}}dx-(\tan x-x)\\\end{aligned}}$

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

${\begin{aligned}I_{22}&=\int t^{2}dt-(\tan x-x)\\&={\frac {1}{3}}t^{3}-\tan x+x+C\\&={\frac {1}{3}}\tan ^{3}x-\tan x+x+C\end{aligned}}$

${\begin{aligned}I_{23}&=\int {\frac {\sin 2x}{1+\sin x}}dx\\&=\int {\frac {2\sin x\cos x}{1+\sin x}}dx\end{aligned}}$

$t=\sin x$ とおくと $dt=\cos xdx$ なので、

${\begin{aligned}I_{23}&=\int {\frac {2t}{1+t}}dt\\&=2\int \left(1-{\frac {1}{1+t}}\right)dt\\&=2t-2\log |1+t|+C\\&=2\sin x-2\log \left(1+\sin x\right)+C\end{aligned}}$

${\begin{aligned}I_{24}&=\int {\frac {x}{1-\cos x}}dx\\&=\int {\frac {x}{2\sin ^{2}{\frac {x}{2}}}}dx\\&=-{\frac {x}{\tan {\frac {x}{2}}}}+\int {\frac {1}{\tan {\frac {x}{2}}}}dx\\&=-{\frac {x}{\tan {\frac {x}{2}}}}+\int {\frac {\cos {\frac {x}{2}}}{\sin {\frac {x}{2}}}}dx\end{aligned}}$

$t=\sin {\frac {x}{2}}$ とおくと $dt={\frac {1}{2}}\cos {\frac {x}{2}}dx$ なので、

${\begin{aligned}I_{24}&=-{\frac {x}{\tan {\frac {x}{2}}}}+2\int {\frac {dt}{t}}\\&=-{\frac {x}{\tan {\frac {x}{2}}}}+2\log |t|+C\\&=-{\frac {x}{\tan {\frac {x}{2}}}}+2\log \left|\sin {\frac {x}{2}}\right|+C\end{aligned}}$

${\begin{aligned}I_{25}&=\int {\frac {1}{1-\sin x}}dx\\&=\int {\frac {1+\sin x}{1-\sin ^{2}x}}dx\\&=\int \left({\frac {1}{\cos ^{2}x}}+{\frac {\sin x}{\cos ^{2}x}}\right)dx\\&=\tan x+\int {\frac {\sin x}{\cos ^{2}x}}dx\end{aligned}}$

$t=\cos x$ とおくと $dt=-\sin xdx$ なので、

${\begin{aligned}I_{25}&=\tan x-\int {\frac {dt}{t^{2}}}\\&=\tan x+{\frac {1}{t}}+C\\&=\tan x+{\frac {1}{\cos x}}+C\end{aligned}}$

${\begin{aligned}I_{26}&=\int {\frac {\sqrt {x}}{{\sqrt[{4}]{x^{3}}}+1}}dx\end{aligned}}$

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

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

${\begin{aligned}I_{27}&=\int \log(x^{2}-1)dx\\&=\int \log(x+1)(x-1)dx\\&=\int \left(\log |x+1|+\log |x-1|\right)dx\\&=(x+1)\log |x+1|-x+(x-1)\log |x-1|-x+C\\&=(x+1)\log |x+1|+(x-1)\log |x-1|-2x+C\end{aligned}}$

${\begin{aligned}I_{28}&=\int {\frac {e^{3x}}{e^{x}-1}}dx\end{aligned}}$

$t=e^{x}$ とおくと $dt=e^{x}dx$ なので、

${\begin{aligned}I_{28}&=\int {\frac {t^{2}}{t-1}}dt\\&=\int \left(t+1+{\frac {1}{t-1}}\right)dt\\&={\frac {1}{2}}t^{2}+t+\log |t-1|+C\\&={\frac {1}{2}}e^{2x}+e^{x}+\log |e^{x}-1|+C\end{aligned}}$

${\begin{aligned}I_{29}&=\int {\frac {e^{x}}{e^{x}-e^{-x}}}dx\\\end{aligned}}$

分子分母に $e^{x}$ を掛けて、

${\begin{aligned}I_{29}&=\int {\frac {e^{2x}}{e^{2x}-1}}dx\end{aligned}}$

$t=e^{2x}$ とおくと $dt=2e^{2x}dx$ なので、

${\begin{aligned}I_{29}&={\frac {1}{2}}\int {\frac {dt}{t-1}}\\&={\frac {1}{2}}\log |t-1|+C\\&={\frac {1}{2}}\log |e^{2x}-1|+C\end{aligned}}$

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

$I_{30}$ について補足
2行目から3行目への変形では、 ${\frac {1}{3x-2-{\sqrt {10}}}}-{\frac {1}{3x-2+{\sqrt {10}}}}={\frac {(3x-2+{\sqrt {10}})-(3x-2-{\sqrt {10}})}{(3x-2-{\sqrt {10}})(3x-2+{\sqrt {10}})}}={\frac {2{\sqrt {10}}}{(3x-2-{\sqrt {10}})(3x-2+{\sqrt {10}})}}$ であることに注意せよ。

3行目から4行目への変形では、 $\left(\log |3x-2-{\sqrt {10}}|\right)'={\frac {3}{3x-2-{\sqrt {10}}}},\ \left(\log |3x-2+{\sqrt {10}}|\right)'={\frac {3}{3x-2+{\sqrt {10}}}}$ であることに注意せよ。

${\begin{aligned}I_{31}&=\int {\frac {dx}{\sqrt {2x^{2}-4x+3}}}\\&={\frac {1}{\sqrt {2}}}\int {\frac {dx}{\sqrt {(x-1)^{2}+{\frac {1}{2}}}}}\\\end{aligned}}$

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

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

${\begin{aligned}I_{32}&=\int {\frac {\tan x}{\cos ^{3}x}}dx\\&=\int {\frac {\sin x}{\cos ^{4}x}}dx\\\end{aligned}}$

$t=\cos x$ とおくと $dt=-\sin xdx$ なので、

${\begin{aligned}I_{32}&=-\int {\frac {dt}{t^{4}}}\\&={\frac {1}{3t^{3}}}+C\\&={\frac {1}{3\cos ^{3}x}}+C\end{aligned}}$

${\begin{aligned}I_{33}&=\int {\frac {dx}{1+\cos x}}\\&=\int {\frac {1-\cos x}{1-\cos ^{2}x}}dx\\&=\int \left({\frac {1}{\sin ^{2}x}}-{\frac {\cos x}{\sin ^{2}x}}\right)dx\\&=-{\frac {1}{\tan x}}-\int {\frac {\cos x}{\sin ^{2}x}}dx\end{aligned}}$

$t=\sin x$ とおくと $dt=\cos xdx$ なので、

${\begin{aligned}I_{33}&=-{\frac {1}{\tan x}}-\int {\frac {dt}{t^{2}}}\\&=-{\frac {1}{\tan x}}+{\frac {1}{t}}+C\\&=-{\frac {1}{\tan x}}+{\frac {1}{\sin x}}+C\end{aligned}}$

別解

${\begin{aligned}I_{33}&=\int {\frac {dx}{1+\cos x}}\\&=\int {\frac {dx}{2\cos ^{2}{\frac {x}{2}}}}\\&=\tan {\frac {x}{2}}+C\end{aligned}}$

${\begin{aligned}I_{34}&=\int x\tan ^{2}xdx\\&=\int \left({\frac {x}{\cos ^{2}x}}-x\right)dx\\&=x\tan x-\int \tan xdx-{\frac {1}{2}}x^{2}\\&=x\tan x+\log |\cos x|-{\frac {1}{2}}x^{2}+C\end{aligned}}$

${\begin{aligned}I_{35}&=\int e^{x}\cos xdx\\&=e^{x}\cos x+\int e^{x}\sin xdx\\&=e^{x}\cos x+e^{x}\sin x-\int e^{x}\cos xdx\\&=e^{x}\cos x+e^{x}\sin x-I_{35}\end{aligned}}$

なので、

${\begin{aligned}2I_{35}&=e^{x}\sin x+e^{x}\cos x+C'\\I_{35}&={\frac {e^{x}}{2}}(\sin x+\cos x)+C\end{aligned}}$

${\begin{aligned}I_{36}&=\int {\sqrt {x^{2}-x-1}}dx\\&=\int {\sqrt {\left(x-{\frac {1}{2}}\right)^{2}-{\frac {5}{4}}}}dx\\\end{aligned}}$

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

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

別解

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

$\left\{t-\left(x-{\frac {1}{2}}\right)\right\}^{2}=\left({\sqrt {\left(x-{\frac {1}{2}}\right)^{2}-{\frac {5}{4}}}}\right)^{2}$

$2t\left(x-{\frac {1}{2}}\right)=t^{2}+{\frac {5}{4}}$

$x-{\frac {1}{2}}={\frac {1}{2}}\left(t+{\frac {5}{4t}}\right)$ …②

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

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

$dx={\frac {1}{2}}\left(1-{\frac {5}{4t^{2}}}\right)dt$ …④

②③④より

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

${\begin{aligned}I_{37}&=\int \sin 2x\cdot \cos 3xdx\\&={\frac {1}{2}}\int (\sin 5x-\sin x)dx\\&=-{\frac {1}{10}}\cos 5x+{\frac {1}{2}}\cos x+C\end{aligned}}$

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

${\begin{aligned}I_{39}&=\int {\frac {1+\cos ^{3}x}{\cos ^{2}x}}dx\\&=\int \left({\frac {1}{\cos ^{2}x}}+\cos x\right)dx\\&=\tan x+\sin x+C\end{aligned}}$

${\begin{aligned}I_{40}&=\int {\frac {dx}{1+\sin x}}\\&=\int {\frac {1-\sin x}{1-\sin ^{2}x}}dx\\&=\int \left({\frac {1}{\cos ^{2}x}}-{\frac {\sin x}{\cos ^{2}x}}\right)dx\\&=\tan x-\int {\frac {\sin x}{\cos ^{2}x}}dx\\\end{aligned}}$

$t=\cos x$ とおくと $dt=-\sin xdx$ なので、

${\begin{aligned}I_{40}&=\tan x+\int {\frac {dt}{t^{2}}}\\&=\tan x-{\frac {1}{t}}+C\\&=\tan x-{\frac {1}{\cos x}}+C\end{aligned}}$

${\begin{aligned}I_{41}&=\int {\frac {dx}{e^{x}+1}}\end{aligned}}$

$t=e^{x}$ とおくと $dt=e^{x}dx$ なので、

${\begin{aligned}I_{41}&=\int {\frac {dt}{t(t+1)}}\\&=\int \left({\frac {1}{t}}-{\frac {1}{t+1}}\right)dt\\&=\log |t|-\log |t+1|+C\\&=x-\log(e^{x}+1)+C\end{aligned}}$

${\begin{aligned}I_{42}&=\int \left(\log x\right)^{2}dx\\&=x\left(\log x\right)^{2}-2\int \log xdx\\&=x\left(\log x\right)^{2}-2x\log x+2x+C\end{aligned}}$

${\begin{aligned}I_{43}&=\int {\frac {\sqrt {1-x^{2}}}{x}}dx\end{aligned}}$

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

${\begin{aligned}I_{43}&=\int {\frac {t^{2}}{t^{2}-1}}dt\\&=\int \left(1+{\frac {1}{t^{2}-1}}\right)dt\\&=\int \left(1+{\frac {1}{2(t-1)}}-{\frac {1}{2(t+1)}}\right)dt\\&=t+\log {\sqrt {\left|{\frac {t-1}{t+1}}\right|}}+C\\&={\sqrt {1-x^{2}}}+\log {\frac {1-{\sqrt {1-x^{2}}}}{|x|}}+C\end{aligned}}$

${\begin{aligned}I_{44}&=\int {\frac {dx}{x{\sqrt {2x-x^{2}}}}}\\\end{aligned}}$

$t={\frac {1}{x}}$ とおくと $dt=-{\frac {1}{x^{2}}}dx$ なので、

${\begin{aligned}I_{44}&=-\int {\frac {dt}{\sqrt {2t-1}}}\\&=-{\sqrt {2t-1}}+C\\&=-{\sqrt {\frac {2-x}{x}}}+C\end{aligned}}$