第9章 9.6

(1)

$\int_{-1}^{1} \frac{dx}{2-x^2} = \int_{-1}^{1} \frac{1}{2\sqrt{2}}\left(\frac{1}{\sqrt{2}-x}+\frac{1}{\sqrt{2}+x}\right)dx = \frac{1}{2\sqrt{2}}\left[-\log|\sqrt{2}-x|+\log|\sqrt{2}+x|\right]_{-1}^{1} = \frac{1}{2\sqrt{2}}\left[\log\frac{|\sqrt{2}+x|}{|\sqrt{2}-x|}\right]_{-1}^{1}$

$= \frac{1}{2\sqrt{2}}\left(\log\frac{|\sqrt{2}+1|}{|\sqrt{2}-1|}-\log\frac{|\sqrt{2}-x|}{|\sqrt{2}+x|}\right) = \frac{1}{2\sqrt{2}}\left(\log\left(\frac{|\sqrt{2}+1|}{|\sqrt{2}-1|}\right)^{2}\right) = \frac{1}{\sqrt{2}}\log\frac{|\sqrt{2}+1|}{|\sqrt{2}-1|} = \frac{1}{\sqrt{2}}\log\frac{(\sqrt{2}+1)(\sqrt{2}+1)}{2-1}$

$= \frac{1}{\sqrt{2}}\log(\sqrt{2}+1)^{2} = \sqrt{2}\log(\sqrt{2}+1)$

(2)

$\int_{1}^{9}\frac{(1-\sqrt{x})^{2}}{x}dx = \int_{1}^{9}\frac{1-2\sqrt{x}+x}{x}dx = \int_{1}^{9}\left(\frac{1}{x}-\frac{2}{\sqrt{x}}+1\right)dx = \left[\log x - 4\sqrt{x}+x\right]_{1}^{9}$

$= \log 9 - 4\sqrt{9}+9-\log 1+4-1 = \log 9 = 2\log 3$

(3)

$\int_{-1}^{1}\sqrt{1+x^{2}}\,dx = \left[x\sqrt{1+x^{2}}\right]_{-1}^{1} - \int_{-1}^{1}\frac{x^{2}}{\sqrt{1+x^{2}}}dx = 2\sqrt{2} - \int_{-1}^{1}\left(\frac{1}{\sqrt{1+x^{2}}}+\frac{1+x^{2}}{\sqrt{1+x^{2}}}\right)dx$

$2\int_{-1}^{1}\sqrt{1+x^{2}}\,dx = 2\sqrt{2} - \int_{-1}^{1}\frac{1}{\sqrt{1+x^{2}}}dx = 2\sqrt{2} - \left[\log\left|x+\sqrt{1+x^{2}}\right|\right]_{-1}^{1} = 2\sqrt{2} - \left(\log\frac{|1+\sqrt{2}|}{|-1+\sqrt{2}|}\right)$

$\int_{-1}^{1}\sqrt{1+x^{2}}\,dx = \sqrt{2} - \frac{1}{2}\log\left(\frac{(1+\sqrt{2})^{2}}{-1+2}\right) = \sqrt{2} - \log(1+\sqrt{2})$

(4)

$\int_{1}^{2}(\log x)^{2}dx = \left[x(\log x)^{2}\right]_{1}^{2} - \int_{1}^{2}x\cdot 2\log x\cdot\frac{1}{x}dx = 2(\log 2)^{2}-2\int_{1}^{2}\log x\,dx$

$= 2(\log 2)^{2}-2\left\{\left[x\cdot\log x\right]_{1}^{2}-\int_{1}^{2}x\cdot\frac{1}{x}dx\right\} = 2(\log 2)^{2}-4\log 2+2\left[x\right]_{1}^{2} = 2(\log 2)^{2}-4\log 2+2 = 2(\log 2-1)^{2}$

(5)

$\int_{0}^{\frac{\pi}{3}}\tan x\,dx = \int_{0}^{\frac{\pi}{3}}\frac{\sin x}{\cos x}dx$

$\cos x = t$ とおくと，$-\sin x\,dx = dt$，$x:0\to\dfrac{\pi}{3}$，$t:1\to\dfrac{1}{2}$ これから

$= \int_{1}^{\frac{1}{2}}-\frac{1}{t}dt = \int_{\frac{1}{2}}^{1}\frac{1}{t}dt = \left[\log t\right]_{\frac{1}{2}}^{1} = -\log\frac{1}{2} = \log 2$

(6)

$\int_4^5 \frac{dx}{x^2-4x+3} = \int_4^5 \frac{dx}{(x-3)(x-1)} = \int_4^5 \frac{1}{2}\left(\frac{1}{x-3}-\frac{1}{x-1}\right)dx = \frac{1}{2}\left[\log\frac{|x-3|}{|x-1|}\right]_4^5$

$= \frac{1}{2}\left(\log\frac{1}{2}-\log\frac{1}{3}\right) = \frac{1}{2}(\log 3-\log 2)$

(7)

$\int_0^1 xe^{x^2}\,dx \qquad x^2=t \text{ とおくと} \qquad 2x\,dx=dt \qquad x:0\to1 \qquad t:0\to1 \qquad \text{これから}$

$= \int_0^1 \frac{1}{2}e^t dx = \frac{1}{2}[e^t]_0^1 = \frac{1}{2}(e-1)$

(8)

$\int_0^1 e^{\sqrt{x}}\,dx \qquad \sqrt{x}=t \text{ とおくと} \qquad x=t^2 \qquad dx=2t\,dt \qquad x:0\to1 \qquad t:0\to1 \qquad \text{これから}$

$= \int_0^1 2te^t dt = 2\left\{[te^t]_0^1 - \int_0^1 e^t dt\right\} = 2\{e-[e^t]_0^1\} = 2(e-e+1) = 2$