\(\displaystyle\fbox{30}***\qquad\int 2^{\log x} dx \) $$\\$$ [導出] $$ \begin{align*} a^{\log_b x}&=\left(b^{\log_b a}\right)^{\log_b x}\\ &=b^{\log_b a\log_b x}\\ &=\left(b^{\log_b x}\right)^{\log_b a}\\ &=x^{\log_b a} \end{align*} $$ \begin{align*} \int 2^{\log x} dx&=\int x^{\log2}dx\\ &=\frac{1}{\log2 +1}x^{\log2 +1}+C\\ &=\frac{x\cdot 2^{\log x}}{\log 2 +1}+C\qquad(答) \end{align*}

\(\displaystyle\fbox{31}**\qquad\int \frac{\log\left(\log x\right)}{x\log x} dx\) $$ \begin{align*} t&=\log x\\ dt&=\frac{1}{x}dx\\ \end{align*} $$ と置換すると, \begin{align*} &\int \frac{\log\left(\log x\right)}{x\log x} dx \\ &=\int\frac{\log t}{t}dt\\ &=\int\left(\log t\right)\cdot\left(\log t\right)'dt\\ &=\frac{1}{2}\left(\log x\right)^2 +C\\ &=\frac{1}{2}\lbrace\log(\log x)\rbrace^2 +C\qquad(答) \end{align*}

\(\displaystyle\fbox{32}**\qquad\int\frac{1}{x\left(\log x\right)^2}dx\) $$ \begin{align*} %Q32 \int\frac{1}{x\left(\log x\right)^2}dx&=\int \left(\log x\right)'\left(\log x\right)^{-2}dx\\ &=-\left(\log x\right)^{-1}+C\\ &=-\frac{1}{\log x}+C\qquad(答)\\ \end{align*} $$

\(\displaystyle\fbox{33}*\qquad\int \sqrt{e^x}dx\) $$ \begin{align*} %Q33 \int \sqrt{e^x}dx&=\int \left(e^x\right)^{\frac{1}{2}}dx\\ &=\int e^{\frac{x}{2}}dx\\ &=2e^{\frac{x}{2}}+C\qquad(答) \end{align*} $$

\(\displaystyle\fbox{34}*****\quad\int^1_{-1}\frac{x^2}{1+e^x}dx\) $$ \begin{align*} %Q34 x=-t,\quad dx&=-dtと置換する.\\ \int^1_{-1}\frac{x^2}{1+e^x}dx&=\int^0_{-1}\frac{x^2}{1+e^x}dx+\int^1_0\frac{x^2}{1+e^x}dx\\ &=\int^1_0\frac{t^2}{1+e^{-t}}dt+\int^1_0\frac{x^2}{1+e^x}dx\\ &=\int^1_0\frac{t^2 e^t}{1+e^{t}}dt+\int^1_0\frac{x^2}{1+e^x}dx\\ &=\int^1_0\frac{x^2\left(1+e^x\right)}{1+e^x}dx\\ &=\int^1_0 x^2dx\\ &=\frac{1}{3}\qquad(答) \end{align*} $$

\(\displaystyle\fbox{35}**\qquad\int^{e^2}_e x^{\frac{1}{\log x}} dx\) $$ \begin{align*} %Q35 y&=x^{\frac{1}{\log x}}とおくと, \\ \log y&=\log x^{\frac{1}{\log x}}\\ &=\frac{1}{\log x}\log x=1\\ \therefore y&=e\\ \int^{e^2}_e x^{\frac{1}{\log x}}dx&=e\int^{e^2}_e dx\\ &=e\left(e^2-e\right)\\ &=e^3-e^2\qquad (答) \end{align*} $$

\(\displaystyle\fbox{36}***\qquad\int\frac{1}{1-e^{-x}}\) $$ \begin{align*} %Q36 \int\frac{1}{1-e^{-x}}&=\int\frac{e^x}{e^x-1}\\ &=\int\frac{\left(e^x-1\right)'}{e^x-1}dx\\ &=\log|e^x-1|+C\qquad(答) \end{align*} $$

\(\displaystyle \fbox{37}****\qquad\int x^x\left(\log x+1\right)dx\) $$ \begin{align*} %Q37 y=x^xとおくと&,\\ \log y&=\log x^x\\ &=x\log x\\ \frac{y'}{y}&=\log x+1\\ y'&=x^x\left(\log x+1\right)\\ \int x^x\left(\log x+1\right)dx&=\int \left(x^x\right)'dx\\ &=x^x+C\qquad(答) \end{align*} $$

\(\displaystyle \fbox{38}***\qquad\int^{\frac{1}{2}\log 3}_0\frac{e^x}{1+e^{2x}}dx\) $$ \begin{align*} t&=e^x\qquad\qquad\\ dt&=e^x dx\qquad \end{align*} $$ \begin{array}{|c||ccc|} %増減表 \hline x & 0 &\rightarrow & \log{\sqrt{3}} \\ \hline t & 1 &\rightarrow &\sqrt{3} \\ \hline \end{array} \begin{align*} t&=\tan{\theta}\qquad\\ dt&=\frac{1}{\cos^2\theta}d\theta\quad \end{align*} \begin{array}{|c||ccc|} %増減表 \hline x & 1&\rightarrow &\sqrt{3} \\ \hline t & \frac{\pi}{4} &\rightarrow &\frac{\pi}{3} \\ \hline \end{array} と置換すると, \begin{align*} \int^{\frac{1}{2}\log 3}_0\frac{e^x}{1+e^{2x}}&=\int^{\frac{1}{2}\log 3}_0\frac{1}{1+\left(e^x\right)^2}e^x dx\\ &=\int^{\sqrt{3}}_1\frac{1}{1+t^2}dt\\ &=\int^{\frac{\pi}{3}}_{\frac{\pi}{4}}d\theta\\ &=\frac{\pi}{12}\qquad(答) \end{align*}

\(\displaystyle \fbox{39}*\qquad\int e^{e^x+x}dx\) $$ \begin{align*} %Q39 \int e^{e^x+x}dx&=\int e^{e^x}\cdot e^x dx\\ &=\int \left(e^{e^x}\right)'dx\\ &=e^{e^x}+C\qquad(答) \end{align*} $$

\(\displaystyle \fbox{40}****\qquad\int^1_0 \log\left(x^2+1\right)dx\) \begin{align*} x&=\tan\theta\quad\\ dx&=\frac{1}{\cos^2\theta} d\theta\quad \end{align*} \begin{array}{|c||ccc|} %増減表 \hline x & 0 & \rightarrow & 1 \\ \hline \theta & 1 & \rightarrow & \frac{\pi}{4} \\ \hline \end{array} と置換すると, \begin{align*} \int^1_0 \log\left(x^2+1\right)dx&= \left[x\log\left(x^2+1\right)\right]^1_0-\int^1_0 x\cdot\frac{2x}{x^2+1}dx\\ &=\log 2-2\int^1_0\frac{x^2}{x^2+1}dx\\ &=\log 2-2\left\lbrace\int^1_0 1dx-\int^1_0\frac{1}{x^2+1}dx\right\rbrace\\ &=\log 2-2+2\int_{0}^{\frac{\pi}{4}}d\theta\\ &=\log 2 -2+\frac{\pi}{2}\qquad(答) \end{align*}

\(\displaystyle \fbox{41}***\qquad\int\frac{\log x}{x^2}dx\) \begin{align*} %Q41 \int\frac{\log x}{x^2}dx&=-\frac{1}{x}\log x+\int\frac{1}{x}\cdot\frac{1}{x}dx\\ &=-\frac{1}{x}\log x-\frac{1}{x}+C\\ &=-\frac{\log x+1}{x}+C\qquad(答) \end{align*}

\(\displaystyle\fbox{42}***\qquad\int\left(\frac{\log x}{x}\right)^2dx \) \begin{align*} %Q42 t=\log x,\quad dt&=\frac{1}{x}dxと置換すると,\\ \int\left(\frac{\log x}{x}\right)^2dx&=\int\frac{t^2}{e^t}dt\\ &=\int t^2 e^{-t}dt\\ &=-t^2e^{-t}+2\int te^{-t}dt\\ &=-t^2e^{-t}+2\left(-te^{-t}+\int e^{-t}dt\right)\\ &=-t^2e^{-t}-2te^{-t}-2e^{-t}+C\\ &=-\frac{\left(\log x\right)^2+2\log x +2}{x}+C\qquad(答) \end{align*}

\(\displaystyle\fbox{43}***\qquad\int\frac{1}{x\left(4-\left(\log x\right)^2\right)}dx \) \begin{align*} %Q43 t=\log x,\quad dt&=\frac{1}{x}dxと置換すると,\\ \int\frac{1}{x\left(4-\left(\log x\right)^2\right)}dx&=\int \frac{1}{4-t^2}dt\\ &=\int \frac{dt}{\left(2-t\right)\left(2+t\right)}\\ &=\frac{1}{4}\int\left(\frac{1}{2-t}+\frac{1}{2+t}\right)dt\\ &=\frac{1}{4}\left(-\log|2-t|+\log |2+t|\right)+C\\ &=\frac{1}{4}\left(-\log|2-\log x|+\log|2+\log x|\right)+C\qquad(答) \end{align*}

\(\displaystyle \fbox{44}**\qquad\int\frac{\left(\log x+3\right)^2}{x}dx\) \begin{align*} %Q44 \int\frac{\left(\log x+3\right)^2}{x}dx&=\int\left(\log x+3\right)'\left(\log x+3\right)^2dx\\ &=\frac{1}{3}\left(\log x+3\right)^3+C\qquad(答) \end{align*}

\(\displaystyle \fbox{45}****\qquad\int^{\sqrt{3}}_1\frac{1}{x^2}\log\sqrt{1+x^2}dx\) \begin{align*} %Q45 \int^{\sqrt{3}}_1&\frac{1}{x^2}\log\sqrt{1+x^2}dx=\frac{1}{2}\int^{\sqrt{3}}_1\left(-\frac{1}{x}\right)'\log\left(1+x^2\right)dx\\ &=\frac{1}{2}\left[\frac{1}{x}\log\left(1+x^2\right)\right]^{\sqrt{3}}_1-\frac{1}{2}\int^{\sqrt{3}}_1\left(-\frac{1}{x}\right)\frac{2x}{1+x^2}dx\\ &=\frac{1}{2}\left(-\frac{1}{\sqrt{3}}\log 4+\log 2\right)+\int_{1}^{\sqrt{3}}\frac{1}{1+x^2}dx\\ \end{align*} ここで \begin{align*} x&=\tan\theta\quad\\ dx&=\frac{1}{\cos^2\theta} d\theta\quad\; \end{align*} \begin{array}{|c||ccc|} %増減表 \hline x & 1 & \rightarrow & \sqrt{3} \\ \hline \theta & \frac{\pi}{4} & \rightarrow & \frac{\pi}{3} \\ \hline \end{array} と置換すると, \begin{align*} &=\left(\frac{1}{2}-\frac{1}{\sqrt{3}}\right)\log 2+\frac{\pi}{12}\qquad(答) \end{align*}

\(\displaystyle \fbox{46}***\qquad\int^2_0\frac{e^x}{e^x+e^{2-x}}dx\) \begin{align*} %Q46 \int^2_0\frac{e^x}{e^x+e^{2-x}}dx&=\int^2_0\frac{e^{2x}}{e^{2x}+e^2}dx\\ &=\frac{1}{2}\int^2_0\frac{\left(e^{2x}+e^2\right)'}{e^{2x}+e^2}dx\\ &=\frac{1}{2}\left[\log \left(e^{2x}+e^2\right)\right]^2_0\\ &=\frac{1}{2}\left\lbrace\log\left(e^4+e^2\right)-\log\left(1+e^2\right)\right\rbrace\\ &=\frac{1}{2}\log\frac{e^2\left(e^2+1\right)}{1+e^2}\\ &=1\qquad(答)\\\\ \end{align*} \begin{align*} \qquad\qquad～King \; Property～\qquad\qquad\qquad\qquad\\ \qquad\qquad\qquad\int^b_a f(x)dx=\int^b_a f(a+b-x)\qquad\qquad\qquad\\ \qquad\qquad\qquad2I=\int^b_a f(x)+f(a+b-x)dx\qquad\qquad\qquad\\ \end{align*} \(King \; Property\)を用いた別解を示す \begin{align*} I&=\int^2_0\frac{e^x}{e^x+e^{2-x}}dx\\ 2I&=\int^2_0\frac{e^x}{e^x+e^{2-x}}+\frac{e^{2-x}}{e^{2-x}+e^x}\\ &=\int^2_0 1dx=2\\ \qquad\qquad\qquad\qquad\therefore I&=1\qquad(答) \end{align*}

\(\displaystyle \fbox{47}**\qquad\int\frac{3^x}{3^x+\log 3}dx\) \begin{align*} %Q47 \int\frac{3^x}{3^x+\log 3}dx&=\int\frac{1}{\log 3}\frac{\left(3^x+\log 3\right)'}{3^x+\log 3}dx\\ &=\frac{1}{\log 3}\log\left(3^x+\log 3\right)+C\qquad(答) \end{align*}

\(\displaystyle \fbox{48}*****\int^4_2\frac{\sqrt{\log(9-x)}}{\sqrt{\log(9-x)}+\sqrt{\log(x+3)}}dx\) \begin{align*} %Q48 I&=\int^4_2\frac{\sqrt{\log(9-x)}}{\sqrt{\log(9-x)}+\sqrt{\log(x+3)}}dx\\ &=\int^4_2\frac{\sqrt{\log(x+3)}}{\sqrt{\log(x+3)}+\sqrt{\log(9-x)}}dx\\ 2I&=\int^4_2 1dx\\ &=2\\ \therefore I&=1 \end{align*}

\(\displaystyle \fbox{49}*****\quad\int^1_0\frac{\log(1+x)}{1+x^2}dx\) \begin{align*} x&=\tan\theta\quad\\ dx&=\frac{1}{\cos^2\theta} d\theta\quad \end{align*} \begin{array}{|c||ccc|} %増減表 \hline x & 0 & \rightarrow & 1 \\ \hline \theta & 0 & \rightarrow & \frac{\pi}{4} \\ \hline \end{array} \begin{align*} \frac{\pi}{4}-\theta&=t\quad\\ d\theta&=-dt\qquad \end{align*} \begin{array}{|c||ccc|} %増減表 \hline \theta & 0 & \rightarrow & \frac{\pi}{4} \\ \hline t & \frac{\pi}{4} & \rightarrow & 0 \\ \hline \end{array} と置換すると, \begin{align*} &=\int^{\frac{\pi}{4}}_0\log(1+\tan\theta)d\theta\\ &=\int^{\frac{\pi}{4}}_0\log(\frac{\cos\theta+\sin\theta}{\cos\theta})d\theta\\ &=\int^{\frac{\pi}{4}}_0\log(\frac{\sqrt{2}\cos(\frac{\pi}{4}-\theta)+\sin\theta}{\cos\theta})d\theta\\ &=\log\sqrt{2}\int^{\frac{\pi}{4}}_0 d\theta+\cancel{\int^{\frac{\pi}{4}}_0 \log\cos(\frac{\pi}{4}-\theta)d\theta}\cancel{-\int^{\frac{\pi}{4}}_0 \log\cos\theta d\theta}\\ &=\frac{\pi}{8}\log 2\qquad(答) \end{align*}

\(\displaystyle \fbox{50}***\qquad\int\frac{\log x}{(x+1)^3}dx \) \begin{align*} %Q50 &=-\frac{1}{2}(x+1)^{-2}\log x+\frac{1}{2}\int\frac{1}{x(x+1)^2}dx\\ &=-\frac{\log x}{2(x+1)^2}+\frac{1}{2}\left\lbrace\log x-\log(x+1)+\frac{1}{x+1}\right\rbrace+C\\ &=\frac{1}{2}\left\lbrace\frac{x(x+2)}{(x+1)^2}\log x-\log(x+1)+\frac{1}{x+1}\right\rbrace +C\quad(答) \end{align*}

\(\displaystyle\fbox{51}**\qquad\int^{e^e}_e\frac{\log x\cdot\log(\log x)}{x}dx\) \begin{align*} t&=\log x\quad\\ dt&=\frac{1}{x} dx \qquad\quad \end{align*} \begin{array}{|c||ccc|} %増減表 \hline x & e & \rightarrow & e^e \\ \hline t & 1 & \rightarrow & e \\ \hline \end{array} と置換すると, \begin{align*} &=\int^e_1 t\cdot\log t dt\\ &=\left[\frac{1}{2}t^2\log t dt\right]^e_1 -\int^e_1\frac{1}{2}t^2\cdot\frac{1}{t}dt\\ &=\frac{1}{2}e^2-\frac{1}{2}\left[\frac{1}{2}t^2\right]^e_1\\ &=\frac{1}{2}e^2-\frac{1}{4}(e^2-1)\\ &=\frac{e^2-1}{4}\qquad(答)\\ \end{align*}

\(\displaystyle \fbox{52}*\qquad\int^2_0 x\log(x+1)dx\) \begin{align*} %Q52 \int^2_0 x\log(x+1)dx=\left[\frac{x^2}{2}\log(x+1)-\int\frac{x^2}{2}\cdot\frac{1}{x+1}dx\right]^2_0\\ &=\left[\frac{x^2}{2}\log(x+1)-\int\frac{1}{2}(\frac{x^2-1+1}{x+1})dx\right]^2_0\\ &=\left[\frac{x^2}{2}\log(x+1)-\int\frac{1}{2}(\frac{(x+1)(x-1)+1}{x+1})dx\right]^2_0\\ &=\left[\frac{x^2}{2}\log(x+1)-\int\frac{1}{2}(x-1+\frac{1}{x+1})dx\right]^2_0\\ &=\left[\frac{x^2}{2}\log(x+1)-\frac{(x-1)^2}{4}-\frac{\log|x+1|}{2}\right]^2_0\\ &=\left(2\log 3-\frac{1}{4}-\frac{\log 3}{2}\right)-\left(0-\frac{1}{4}-0\right)\\ &=\frac{3\log 3}{2}\qquad(答) \end{align*}

\(\displaystyle \fbox{53}**\qquad\int^1_0 x\log(x^2+1)dx\) \begin{align*} u&=x^2+1\qquad\\ du&=2xdx\qquad \end{align*} \begin{array}{|c||ccc|} %増減表 \hline x &0 & \rightarrow & 1 \\ \hline u & 1 & \rightarrow & 2 \\ \hline \end{array} と置換すると, \begin{align*} \int^1_0 x\log(x^2+1)dx&=\int^2_1\frac{1}{2}\log udu\\ &=\frac{1}{2}\left[u(\log u-1)\right]^2_1\\ &=\frac{1}{2}(2\log 2-2+1)\\ &=\frac{1}{2}(2\log 2-1)\qquad(答) \end{align*}

\(\displaystyle \fbox{54}*\qquad\int^e_1\frac{\log x}{x^3}dx\) \begin{align*} %Q54 \int^e_1\frac{\log x}{x^3}dx&=\int^e_1\log x\cdot x^{-3}dx\\ &=\int^e_1\log x\cdot\left(\frac{x^{-2}}{-2}\right)'dx\\ &=\left[\log x\cdot\left(\frac{x^{-2}}{-2}\right)-\int\frac{1}{x}\cdot\left(\frac{x^{-2}}{-2}\right)dx\right]^e_1\\ &=\left[-\frac{1}{2x^2}\log x-\int\frac{x^{-3}}{-2}dx\right]^e_1\\ &=\left[-\frac{1}{2x^2}\log x-\frac{1}{4x^2}\right]^e_1\\ &=-\frac{1}{2e^2}-\frac{1}{4e^2}+\frac{1}{4}\\ &=-\frac{3}{4e^2}+\frac{1}{4}\qquad(答) \end{align*}

\(\displaystyle \fbox{55}**\qquad\int\frac{1}{e^x(e^x+1)}dx\) \begin{align*} %Q55 \int\frac{1}{e^x(e^x+1)}dx&=\int\left(\frac{1}{e^x}-\frac{1}{e^x+1}\right)dx\\ &=-e^{-x}-\int\frac{1}{e^x+1}dx \end{align*} ここで \begin{align*} \quad t&=e^x\quad\\ \quad dt&=e^x dx\quad\\ \end{align*} と置換すると, \begin{align*} &=-e^{-x}-\int\frac{1}{t+1}\cdot \frac{dt}{t}\\ &=-e^{-x}-\int\left(\frac{1}{t}-\frac{1}{t+1}\right)dt\\ &=-e^{-x}-\log|t|+\log|t+1|+C\\ &=-e^{-x}-\log e^x+\log(e^x+1)+C\\ &=-e^{-x}-x+\log(e^x+1)+C\quad(答) \end{align*}