\(\displaystyle\fbox{1}\qquad I=\int \frac{x^3}{(x+1)(x+2)} dx\) $$ \begin{align*} x^3&=(x^2+3x+2)(x-3)+7x+6\\ \end{align*} $$より $$ \begin{align*} \frac{x^3}{(x+1)(x+2)}&=x-3+\frac{7x+6}{(x+1)(x+2)}\\ \end{align*} $$ から $$ \begin{align*} \therefore I&=\int\left(x-3+\frac{-1}{x+1}+\frac{8}{x+2}\right)dx\\ &=\frac{x^2}{2}-3x-\log|x+1|+8\log|x+2| + C\qquad(答)\\ \end{align*} $$

\(\displaystyle\fbox{2}\qquad\int_{0}^{2\pi} \sqrt{1+\cos x} dx\) $$ \begin{align*} %問2 \sqrt{1+\cos x}&=\sqrt{2\cos^2\frac{x}{2}}\\ &=\sqrt{2}|\cos\frac{x}{2}|\\ &=\begin{cases} \sqrt{2}\cos\frac{x}{2} & (0\leq x \leq \pi)\\ -\sqrt{2}\cos\frac{x}{2} & (\pi \leq x \leq 2\pi) \end{cases} \end{align*} $$ $$ \begin{align*} \int_{0}^{2\pi} \sqrt{1+\cos x} dx &=\int_{0}^{\pi} \sqrt{1+\cos x} dx +\int_{\pi}^{2\pi} -\sqrt{1+\cos x} dx\\ &=\sqrt{2}\left\lbrace \left[2\sin\frac{x}{2}\right]^{\pi}_0 -\left[2\sin\frac{x}{2}\right]^{2\pi}_{\pi}\right\rbrace\\ &=\sqrt{2}{(2-0)-(0-2)}\\ &=4\sqrt{2}\qquad(答) \end{align*} $$

\(\displaystyle\fbox{3}\qquad\int^1_0 \sqrt{\frac{1-x}{1+x}}dx\) $$ \begin{flalign*} t^2&=\frac{1-x}{1+x}\\ x(1+x^2)&=1-t^2\\ x&=\frac{1-t^2}{1+t^2}\\ \frac{dx}{dt}&=-\frac{4t^2}{1+t^2}\\ \\ \end{flalign*} $$ 以下のように置換すると$$ \begin{align*} t&=\sqrt{\frac{1-x}{1+x}} \qquad \end{align*} $$$$ \begin{array}{|c||ccc|} %増減表 \hline x & 0 & \rightarrow & 1 \\ \hline t & 1 & \rightarrow & 0 \\ \hline \end{array} $$$$ \begin{align*} \frac{dx}{dt}&=-\frac{4t^2}{1+t^2}\qquad\\ \end{align*} $$ とかけ, $$ \begin{align*} \int^1_0 \sqrt{\frac{1-x}{1+x}}dx&=\int^0_1t\left\lbrace-\frac{4t}{(1+t^2)^2}dt\right\rbrace\\ &=4\int^1_0\frac{t^2}{(1+t^2)^2}dt \end{align*} $$ 以下のように置換すると $$ \begin{align*} t&=\tan\theta \end{align*} $$$$ \begin{array}{|c||ccc|} %増減表 \hline t & 0 & \rightarrow & 1 \\ \hline \theta & 0 & \rightarrow & \frac{\pi}{4} \\ \hline \end{array} $$$$ \begin{align*} dt=\frac{1}{\cos^2\theta}d\theta \qquad \\ \end{align*} $$ とかけ, $$ \begin{align*} &=4\int_{0}^{\frac{\pi}{4}}\tan^2\theta\cdot \cos^4\theta\cdot \frac{1}{\cos^2\theta}d\theta\\ &=\int_{0}^{\frac{\pi}{4}}\sin^2\theta d\theta\\ &=\int_{0}^{\frac{\pi}{4}}\frac{1-\cos2\theta}{2}d\theta\\ &=\frac{1}{2}\left[\theta-\frac{1}{2}\sin2\theta\right]^{\frac{\pi}{4}}_0\\ &=\frac{{\pi}}{8}-\frac{1}{4} \qquad(答)\\ \\ \end{align*} $$

\(\displaystyle\fbox{4}\qquad\int x\sqrt{x+1} dx\) $$ \begin{flalign*} x\sqrt{x+1}&=(x+1)\sqrt{x+1}-\sqrt{x+1}\\ &=(x+1)^{\frac{3}{2}}-(x+1)^{\frac{1}{2}}\\ \int x\sqrt{x+1} dx&=\int (x+1)^{\frac{3}{2}}-(x+1)^{\frac{1}{2}}dx\\ &=\frac{2}{5}(x+1)^{\frac{5}{2}}-\frac{2}{3}(x+1)^{\frac{3}{2}} \qquad(答) \end{flalign*} $$

\(\displaystyle\fbox{5}\qquad\int \frac{1}{x^2}(1+\frac{2}{x})^2 dx\) $$ \begin{align*} %Q5 \left\lbrace1+\frac{2}{x}\right\rbrace'&=-\frac{2}{x^2}より\\ \int -\frac{1}{2}(1+\frac{2}{x})'(1+\frac{2}{x})^2 dx&=-\frac{1}{6}(1+\frac{2}{x})^3+C \qquad(答)\\ \end{align*} $$

\(\displaystyle\fbox{6}\qquad\int^1_0 \sqrt{1+2\sqrt{x}} dx\) $$ \begin{flalign*} t&=\sqrt{1+2\sqrt{x}}\\ \end{flalign*} $$ と置換し, $$ \begin{flalign*} x&=(\frac{t^2-1}{2})^2\\ &=\frac{1}{4}(t^2-1)^2\\ \end{flalign*} $$ となり $$ \begin{align*} x&=\frac{1}{4}(t^2-1)^2\qquad \end{align*} $$ と置換すると, $$ \begin{array}{|c||ccc|} %増減表 \hline x & 0 & \rightarrow & 1 \\ \hline t & 1 & \rightarrow & \sqrt{3} \\ \hline \end{array} $$ $$ \begin{align*} dx=t(t^2-1)dt \qquad\\ \end{align*} $$ とかけ, $$ \begin{align*} \int^1_0 \sqrt{1+2\sqrt{x}} dx&=\int^{\sqrt{3}}_1t\cdot t(t^2-1)dt\\ &=\left[\frac{1}{5}t^5-\frac{1}{3}t^3\right]^{\sqrt{3}}_1\\ &=\frac{4}{5}\sqrt{3}+\frac{2}{15}\qquad(答)\\ \end{align*} $$

\(\displaystyle\fbox{7}* \qquad\int_{1}^{2}\frac{1}{2^x}dx\) $$\\$$ 対数微分法より $$ \begin{align*} y&=2^{-x}\\ \log y&=-x\log 2\\ \frac{y'}{y}&=-\log 2\qquad\\ y'&=-2^{-x}\log 2\\ \end{align*} $$ $$ \begin{align*} (2^{-x})'=-2^{-x}\log 2\\ 2^{-x}=(-\frac{2^{-x}}{\log 2})'\\ \end{align*} $$ とかけるから, $$ \begin{align*} \int_{1}^{2}\frac{1}{2^x}dx=\left[-\frac{2^{-x}}{\log2}\right]^2_1&=-\frac{1}{\log2}(2^{-2}-2^{-1})\\ &=\frac{1}{4\log2}\qquad(答) \end{align*} $$

\(\displaystyle\fbox{8}*\qquad\int^1_0\frac{1}{x^2-2x+4}\) $$ \begin{align*} \int^1_0\frac{1}{x^2-2x+4}=\int^1_0\frac{1}{(x-1)^2 +3}\\ \end{align*} $$ $$ \begin{align*} &x-1=\sqrt{3}\tan\theta\\ &dx=\sqrt{3}\frac{1}{\cos^2\theta} d\theta\; \end{align*} $$ と置換すると, $$ \begin{array}{|c||ccc|} %増減表 \hline x & 0 & \rightarrow & 1 \\ \hline \theta & -\frac{\pi}{6} & \rightarrow & 0 \\ \hline \end{array} $$ $$ \begin{align*} &=\int^0_{-\frac{\pi}{6}}\frac{\cos^2\theta}{3}\cdot \sqrt{3}\cdot \frac{1}{\cos^2\theta}d\theta\\ &=\frac{\sqrt{3}}{3}\cdot\frac{\pi}{6}\\ &=\frac{\sqrt{3}}{18}\pi\qquad(答) \end{align*} $$

\(\displaystyle\fbox{9}****\qquad\int \frac{1}{\sqrt{1+x^2}}dx\) $$ \begin{align*} x&=\frac{e^t-e^{-t}}{2}\\ t&=\log \left(x+\sqrt{x^2+1}\right)\\ dx&=\frac{e^t+e^{-t}}{2}dt\\ \end{align*} $$ とおくと, $$ \begin{align*} 2x=e^t-e^{-t}\\ (e^t)^2-2xe^t-1=0\\ e^t=x+\sqrt{x^2+1}\\ \end{align*} $$ $$ \begin{align*} \int \frac{1}{\sqrt{1+x^2}}dx&=\int\frac{1}{\sqrt{1+\frac{\left(e^t-e^{-t}\right)^2}{4}}}\frac{e^t+e^{-t}}{2}dt\\ &=\int\frac{1}{\frac{e^t+e^{-t}}{2}}\cdot\frac{e^t+e^{-t}}{2}dt\\ &=\int dt\\ &=t+C\\ &=\log\left(x+\sqrt{x^2+1}\right)+C\qquad(答) \end{align*} $$

\(\displaystyle\fbox{10}**\qquad\int x\sqrt{3x-5}dx\) $$ \begin{align*} t&=\sqrt{3x-5}\qquad\\ x&=\frac{t^2+5}{3}\qquad\\ dx&=\frac{2}{3}tdt\qquad\\ \end{align*} $$ と置換すると, $$ \begin{align*} \int \frac{t^2+5}{3}\cdot t\cdot\frac{2}{3}tdt&=\frac{2}{9}\int t^4+5t^2dt\\ &=\frac{2}{9}\left(\frac{1}{5}t^5+\frac{5}{3}t^3\right)+C\\ &=\frac{2}{45}(3x-5)\sqrt{3x-5}(3x+\frac{10}{3})+C\qquad(答) \end{align*} $$

\(\displaystyle\fbox{11}*\qquad\int^2_0 \frac{x}{4+x^2}dx\) $$ \begin{align*} %Q11 u=4+x^2,\quad du&=2xdxと置換\\ I&=\int^5_4\frac{\frac{1}{2}du}{u}\\ &=\left[\frac{1}{2}\log|u|\right]^5_4\\ &=\frac{1}{2}(\log 5-\log 4)\qquad(答) \end{align*} $$

\(\displaystyle\fbox{12}***\qquad\int\frac{x}{(x^2+2)(x^2+3)}dx\) $$ \begin{align*} %Q12 \int\frac{x}{(x^2+2)(x^2+3)}dx&=\int x\left(\frac{1}{x^2+2}-\frac{1}{x^2+3}\right)dx\\ &=\int\left(\frac{x}{x^2+2}-\frac{x}{x^2+3}\right)dx\\ &=\int\left\lbrace\frac{1}{2}\cdot\frac{(x^2+2)'}{x^2+2}-\frac{1}{2}\frac{(x^2+3)'}{x^2+3}\right\rbrace dx\\ &=\frac{1}{2}\log\frac{x^2+2}{x^2+3}+C\qquad(答) \end{align*} $$

\(\displaystyle\fbox{13}*\qquad\int^1_{-1}\frac{x^3}{1+x^2}dx\) $$ \begin{align*} %Q13 \frac{x^3}{1+x^2}は奇関数である為,\\ \int^1_{-1}\frac{x^3}{1+x^2}dx=0\qquad(答) \end{align*} $$

\(\displaystyle\fbox{14}***\qquad\int\frac{4(3+3x-x^2)}{(x-1)^2(x+1)}dx\) $$ \begin{align*} %Q14 \int\frac{4(3+3x-x^2)}{(x-1)^2(x+1)}dx&=\int\left\lbrace-\frac{3}{x-1}+\frac{10}{(x-1)^2}-\frac{1}{x+1}\right\rbrace dx\\ &=\log|\frac{1}{(x-1)^3(x+1)}-\frac{10}{x-1}+C\qquad(答) \end{align*} $$

\(\displaystyle\fbox{15}****\quad\int^1_0\sqrt{\frac{x}{1+x}}dx\) $$ \begin{align*} x&=\tan^2\theta\quad\\ dx&=\frac{2\tan x}{\cos^2\theta}d\theta\;\; \end{align*} $$ $$ \begin{array}{|c||ccc|} %増減表 \hline x & 0 & \rightarrow & 1 \\ \hline \theta & 0 & \rightarrow & \frac{\pi}{4} \\ \hline \end{array} $$ と置換すると, $$ \begin{align*} \int^1_0\sqrt{\frac{x}{1+x}}dx&=\int^{\frac{\pi}{4}}_0\sqrt{\frac{\tan^2\theta}{1+\tan^2\theta}}\cdot 2\tan\theta\frac{1}{\cos^2\theta}d\theta\\ &=\int^{\frac{\pi}{4}}_0 2\tan\theta\cdot\cos\theta\cdot\frac{\sin\theta}{\cos^3\theta}d\theta\\ &=\int^{\frac{\pi}{4}}_0\frac{2\sin^2\theta}{\cos^3\theta}d\theta\\ &=\int^{\frac{\pi}{4}}_0\sin\theta\cdot\left(\frac{1}{\cos^2\theta}\right)'d\theta\\ &=\left[\sin\theta\cdot\frac{1}{\cos^2\theta}\right]^{\frac{\pi}{4}_0}-\int^{\frac{\pi}{4}}_0\frac{1}{\cos^2\theta}d\theta\\ &=\sqrt{2}-\int^{\frac{\pi}{4}}_0 \frac{\cos\theta}{\cos^2\theta}d\theta\\ &=\sqrt{2}-\int^{\frac{\pi}{4}}_0 \frac{\cos\theta}{1-\sin^2\theta}d\theta\\ \end{align*} $$ ここで $$ \begin{align*} t&=\sin\theta\quad\\ dt&=\cos\theta d\theta\qquad \end{align*} $$ $$ \begin{array}{|c||ccc|} %増減表 \hline x & 0 & \rightarrow & 1 \\ \hline \theta & 0 & \rightarrow & \frac{\pi}{4} \\ \hline \end{array} $$ と置換すると, $$ \begin{align*} &=\sqrt{2}-\int^{\frac{1}{\sqrt{2}}}_0 \frac{1}{1-t^2}dt\\ &=\sqrt{2}-\int^{\frac{1}{\sqrt{2}}}_0 \left(\frac{1}{1+t}+\frac{1}{1-t}\right)dt\\ &=\sqrt{2}-\frac{1}{2}\left[\log|1+t|-\log|1-t|\right]^{\frac{1}{\sqrt{2}}}_0\\ &=\sqrt{2}-\frac{1}{2}\log\frac{\sqrt{2}+1}{\sqrt{2}-1}\\ &=\sqrt{2}-\frac{1}{2}\log(\sqrt{2}+1)^2\\ &=\sqrt{2}-\log(\sqrt{2}+1)\qquad(答)\\ \end{align*} $$

\(\displaystyle\fbox{16}***\quad\int\sqrt{x\sqrt{x\sqrt{x\cdots}}}dx\) $$ \begin{align*} %Q16 \sqrt{x\sqrt{x\sqrt{x\cdots}}}&=(x(x(x\cdots)^{\frac{1}{2}})^{\frac{1}{2}})^{\frac{1}{2}}\\ &=x^{\frac{1}{2}}x^{\frac{1}{4}}x^{\frac{1}{8}}\cdots\\ &=x^{\sum\limits^{\infty}_{n=1}\frac{1}{2^n}}=x\\ \therefore\int\sqrt{x\sqrt{x\sqrt{x\cdots}}}dx&=\int xdx =\frac{1}{2}x^2+C\qquad(答) \end{align*} $$

\(\displaystyle\fbox{17}***\qquad\int^1_{\frac{1}{2}}x\sqrt{\frac{1}{x}-1}dx\) $$ \begin{align*} x-\frac{1}{2}&=\frac{1}{2}\sin\theta\\ dx&=\frac{1}{2}\cos\theta d\theta \end{align*} $$ $$ \begin{array}{|c||ccc|} %増減表 \hline x & \frac{1}{2} & \rightarrow & 1 \\ \hline \theta & 0 & \rightarrow & \frac{\pi}{2} \\ \hline \end{array} $$ と置換すると, $$ \begin{align*} \int^1_{\frac{1}{2}}&x\sqrt{\frac{1}{x}-1}dx\\ &=\int^1_{\frac{1}{2}}\sqrt{x-x^2}dx\\ &=\int^1_{\frac{1}{2}}\sqrt{\frac{1}{4}-(x-\frac{1}{2})^2}dx\\ &=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\sqrt{1-\sin^2\theta}\cdot\frac{1}{2}\cos\theta d\theta\\ &=\frac{1}{4}\int_{0}^{\frac{\pi}{2}}\cos^2\theta d\theta\\ &=\frac{1}{4}\int_{0}^{\frac{\pi}{2}}\frac{1+\cos 2\theta}{2}d\theta\\ &=\frac{1}{4}\left[\frac{\theta}{2}+\frac{\sin 2\theta}{4}\right]^{\frac{\pi}{2}}_0\\ &=\frac{\pi}{16}\qquad(答)\\ \end{align*} $$

\(\displaystyle\fbox{18}**\qquad\int\frac{1}{x(x+1)(x+2)}dx\) $$ \begin{align*} %Q18 \int\frac{1}{x(x+1)(x+2)}dx&=\int\left(\frac{1}{2x}-\frac{1}{x+1}+\frac{1}{2(x+2)}\right)dx\\ &=\frac{1}{2}\log|x|-\log|x+1|+\frac{1}{2}\log|x+2|+C\qquad(答) \end{align*} $$

\(\displaystyle\fbox{19}**\qquad\int\frac{x}{1-x^2}dx\) $$ \begin{align*} %Q19 \int\frac{x}{1-x^2}dx&=-\frac{1}{2}\int\frac{(1-x^2)'}{1-x^2}dx\\ &=-\frac{1}{2}\log|1-x^2|+C\qquad(答) \end{align*} $$

\(\displaystyle\fbox{20}***\qquad\int^2_{\sqrt{2}}\frac{1}{x\sqrt{x^2-1}}dx\) $$ \begin{align*} t&=\sqrt{x^2-1}\\ dt&=\frac{x}{\sqrt{x^2-1}}dx\;\; \end{align*} $$ $$ \begin{array}{|c||ccc|} %増減表 \hline x & \sqrt{2} & \rightarrow &2 \\ \hline t & 1 & \rightarrow & \sqrt{3} \\ \hline \end{array} $$ $$ \begin{align*} t&=\tan\theta\\ dt&=\frac{1}{\cos^2\theta}d\theta\quad\;\;\; \end{align*} $$ $$ \begin{array}{|c||ccc|} %増減表 \hline t & 1 & \rightarrow & \sqrt{3} \\ \hline \theta & \frac{\pi}{4} & \rightarrow & \frac{\pi}{3} \\ \hline \end{array} $$ と置換して $$ \begin{align*} \int^2_{\sqrt{2}}\frac{1}{x\sqrt{x^2-1}}dx&=\int^{\sqrt{3}}_1\frac{1}{t^2+1}dt\\ &=\int^{\frac{\pi}{3}}_{\frac{\pi}{4}}d\theta\\ &=\frac{\pi}{12}\qquad(答)\\ \end{align*} $$

\(\displaystyle\fbox{21}**\qquad\int\frac{1}{x\sqrt{1-x^2}}dx\) $$ \begin{align*} t&=\sqrt{1-x^2}\\ dt&=-\frac{x}{\sqrt{1-x^2}}dx\\ \end{align*} $$ と置換すると, $$ \begin{align*} \int\frac{1}{x\sqrt{1-x^2}}dx&=\int\frac{1}{t^2-1}dt\\ &=\frac{1}{2}\int\left(\frac{1}{t-1}-\frac{1}{t+1}\right)dt\\ &=\frac{1}{2}\log|t-1|-\frac{1}{2}\log|t+1|+C\\ &=\frac{1}{2}\log|\sqrt{1-x^2}-1|\\ &-\frac{1}{2}\log|\sqrt{1-x^2}+1|+C\qquad(答)\\ \end{align*} $$

\(\displaystyle\fbox{22}*\qquad\int_{0}^{\pi}\left|\cos x-\cos2x\right|dx\) $$ \begin{align*} %Q22 f(x)&=\cos x-\cos2xとおいて符号を調べる\\ &=\cos x-(2\cos^2x-1)\\ &=-2\cos^2x+\cos x+1\\ &=(1-\cos x)(2\cos x+1)\\ 積分区&間である 0\leq x \leq\pi の時, 1-\cos x\geq0が成立するので\\ 2\cos x&+1=0となる x=\frac{2}{3}\pi の前後で正から負に変わる \end{align*} $$ $$ \begin{align*} \int_{0}^{\pi}\left|f(x)\right|dx&=\int_{0}^{\frac{2}{3}\pi}f(x)dx+\int_{\frac{2}{3}\pi}^{\pi}-f(x)dx\\ &=\int_{0}^{\frac{2}{3}\pi}f(x)dx+\int_{\pi}^{\frac{2}{3}\pi}f(x)dx\\ &=\left[\sin x-\frac{1}{2}\sin2x\right]^{\frac{2}{3}\pi}_0+\left[\sin x-\frac{1}{2}\sin2x\right]^{\frac{2}{3}\pi}_{\pi}\\ &=2\left\lbrace\frac{\sqrt{3}}{2}-\frac{1}{2}\left(-\frac{\sqrt{3}}{2}\right)\right\rbrace\\ &=\frac{3}{2}\sqrt{3}\qquad(答) \end{align*} $$

\(\displaystyle\fbox{23}**\qquad\int_{1}^{e}\sqrt{x}\log xdx\) $$ \begin{align*} %23 \int_{1}^{e}x^{\frac{1}{2}}\log xdx&=\left[\frac{2}{3}x^{\frac{3}{2}}\log x\right]-\int_{1}^{e}\frac{2}{3}x^{\frac{3}{2}}\cdot x^{-1}dx\\ \qquad&=\frac{2}{3}e^{\frac{3}{2}}-\frac{2}{3}\int_{1}^{e}x^{\frac{1}{2}}dx\\ \qquad&=\frac{2}{9}e\sqrt{e}+\frac{4}{9}\qquad(答) \end{align*} $$