解析入門Ⅰ 第Ⅳ章積分法 §５一変数函数の積分　大問3 問題の解法

間違いがあればコメントで教えていただければ幸いです。

$${\bf 3)}$$次の函数の原始函数を求めよ。

$${(\mathrm{i})\:x^{\alpha}\log{x}\;\;(\alpha\not=-1)}$$

$$
\displaystyle\begin{aligned}\int x^{\alpha}\log{x}\:dx&=\int\left(\dfrac{1}{\alpha+1}x^{\alpha+1}\right)^{\prime}\log{x}\:dx\\&=\dfrac{1}{\alpha+1}x^{\alpha+1}\log{x}-\dfrac{1}{\alpha+1}\int x^{\alpha+1}\!\times\dfrac{1}{x}dx\\&=\dfrac{1}{\alpha+1}x^{\alpha+1}\log{x}-\dfrac{1}{(\alpha+1)^{2}}x^{\alpha+1}+C\\&(積分定数をCとする)\end{aligned}
$$

$${(\mathrm{ii})\:x^{n}e^{x}(n\in\mathbb N)}$$

$$
\displaystyle\begin{aligned}\int x^{n}e^{x}dx&=\int x^{n}\left(e^{x}\right)^{\prime}dx\\&=x^{n}e^{x}-\int n\,x^{n-1}e^{x}dx\\&=x^{n}e^{x}-n\,x^{n-1}e^{x}+\int n(n-1)x^{n-2}e^{x}dx\\&\hspace{30mm}\vdots\\&=e^{x}\left(x^{n}-nx^{n-1}+\cdots+(-1)^{n}n!\right)+C\\&=e^{x}\sum^{n}_{k=0}(-1)^{k}\dfrac{n!}{(n-k)!}x^{n-k}+C\end{aligned}
$$

$${(\mathrm{iii})\:(\log{x})^{n}(n\in\mathbb N)}$$

$$
\displaystyle\begin{aligned}\int(\log{x})^{n}dx&=\int(x)^{\prime}(\log{x})^{n}dx\\&=x(\log{x})^{n}-\int{x}(\log{x})^{n-1}n\:\dfrac{1}{x}\;dx\\&=x(\log{x})^{n}\!-\! nx(\log{x})^{n-1}\!+n\int{x}(\log{x})^{n-2}(n-1)\:\dfrac{1}{x}dx\\&=x\left((\log{x})^{n}-n(\log{x})^{n-1}+\cdots+(-1)^{n}n!\right)+C\\&=x\sum^{n}_{k=0}(-1)^{k}\dfrac{n!}{(n-k)!}(\log{x})^{n-k}+C\end{aligned}
$$

$${(\mathrm{iv})\:\arcsin {x}}$$

$$
\displaystyle\begin{aligned}\int \arcsin{x}\;dx&=\int(x)^{\prime}\arcsin{x}\;dx\\&=x\arcsin{x}-\int x\,\dfrac{1}{\sqrt{1-x^{2}}}\:dx\\&=x\arcsin{x}+\sqrt{1-x^{2}}+C\end{aligned}
$$

$${(\mathrm{v})\:\cos^{2}x}$$

$$
\displaystyle\int\cos^{2}x\:dx=\int\dfrac{1+\cos{2x}}{2}\:dx=\dfrac{1}{2}x+\dfrac{1}{4}\sin{2x}+C
$$

$${(\mathrm{vi})\:\dfrac{(\log{x})^{2}}{x}}$$

$$
\displaystyle\int\dfrac{(\log{x})^{2}}{x}\:dx=\int(\log{x})^{\prime}(\log{x})^{2}\:dx=\dfrac{1}{3}(\log{x})^{3}+C
$$

$${(\mathrm{vii})\:\dfrac{(x^{2}+2)}{(x+1)^{3}}}$$

$$
\displaystyle\begin{aligned}\int\dfrac{(x^{2}+2)}{(x+1)^{3}}\:dx&=\int\dfrac{(x+1)^{2}-2(x+1)+3}{(x+1)^{3}}\:dx\\&=\log|x+1|+\dfrac{2}{x+1}-\dfrac{3}{2(x+1)^{2}}+C\end{aligned}
$$

$${(\mathrm{viii})\:\sin^{3}x\cos^{2}x}$$

$$
\displaystyle\begin{aligned}\int\sin^{3}x\cos^{2}x\:dx&=\int\sin{x}(1-\cos^{2}x)\cos^{2}x\:dx\\&=\int(\cos x)^{\prime}(\cos^{2}x-\cos^{4}x)\:dx\\&=-\dfrac{1}{3}\cos^{3}x+\dfrac{1}{5}\cos^{5}x+C\end{aligned}
$$

$${(\mathrm{ix})\:\dfrac{1}{(\sqrt{x}-\sqrt[3]{x})}}$$

$$
\displaystyle\begin{aligned}&\int\dfrac{1}{(\sqrt{x}-\sqrt[3]{x})}\:dx\\&ここで、x=t^{6}と置換すると、\dfrac{dx}{dt}=6t^{5}\end{aligned}\\\begin{aligned}\therefore\int\dfrac{1}{(\sqrt{x}-\sqrt[3]{x})}\:dx&=\int\dfrac{1}{(t^{3}-t^{2})}\:6t^{5}\,dt\\&=6\int\dfrac{t^{3}}{t-1}dt\\&=6\int\dfrac{t^{2}(t-1)+t(t-1)+(t-1)+1}{t-1}dt\\&=2t^{3}+3t^{2}+6t+6\log|t-1|+C\\&=2\sqrt{x}+3\sqrt[3]{x}+6\sqrt[6]{x}+6\log|\sqrt[6]{x}-1|+C\end{aligned}
$$

$${(\mathrm{x})\:e^{ax}\cos bx,(a^{2}\!+b^{2}\not=0)}$$

$$
\displaystyle\int e^{ax}\cos bx\:dxをIとする。\\\begin{aligned}I=\int e^{ax}\cos bx\:dx&=\int\left(\dfrac{1}{a}e^{ax}\right)^{\prime}\cos bx\:dx\\&=\dfrac{1}{a}e^{ax}\cos bx-\dfrac{1}{a}\int e^{ax}b(-\sin bx)\:dx\\&=\dfrac{1}{a}e^{ax}\cos bx+\dfrac{b}{a}\int\left(\dfrac{1}{a}e^{ax}\right)^{\prime}\sin bx\:dx\\&=\dfrac{1}{a}e^{ax}\cos bx+\dfrac{b}{a^{2}}e^{ax}\sin bx-\dfrac{b}{a^{2}}\int e^{ax}b\cos bx\:dx\\&=\dfrac{1}{a}e^{ax}\cos bx+\dfrac{b}{a^{2}}e^{ax}\sin bx-\dfrac{b^{2}}{a^{2}}I\\よって、\dfrac{a^{2}+b^{2}}{a^{2}}I&=\dfrac{1}{a}e^{ax}\cos bx+\dfrac{b}{a^{2}}e^{ax}\sin bx\\\therefore\int e^{ax}\cos bx\:dx&=\dfrac{a}{a^{2}+b^{2}}e^{ax}\cos bx+\dfrac{b}{a^{2}+b^{2}}e^{ax}\sin bx+C\end{aligned}
$$