第5章 5.25

(1)

$\int_0^1 \frac{2x-1}{x(x-1)}dx = \int_0^1 \frac{2x-1}{x^2-x}dx = \left[\log|x(x-1)|\right]_0^1 = \text{発散}$

(2)

$\int_0^\infty \frac{x}{(x^2+4)^2}dx = \left[-\frac{1}{2}\cdot\frac{1}{(x^2+4)}\right]_0^\infty = \frac{1}{8}$

(3)

$\int_0^\infty \frac{x}{(x^2+1)(x^2+4)}dx = \int_0^\infty \left(\frac{x}{x^2+1}-\frac{x}{x^2+4}\right)\frac{1}{3}dx = \frac{1}{3}\left[\frac{1}{2}\log(x^2+1)-\frac{1}{2}\log(x^2+4)\right]_0^\infty$

$=\frac{1}{6}\left[\log\frac{x^2+1}{x^2+4}\right]_0^\infty = \frac{1}{6}\left(\log 1 - \log\frac{1}{4}\right) = \frac{1}{3}\log 2$

(4)

$\int_0^\infty \frac{dx}{(2x+1)^4} = \left[-\frac{1}{3}\cdot\frac{1}{2}\cdot\frac{1}{(2x+1)^3}\right]_0^\infty = \frac{1}{6}$

(5)

$\int_1^\infty \frac{dx}{x+x^2} = \int_1^\infty \frac{1}{x(x+1)}dx = \int_1^\infty \left(\frac{1}{x}-\frac{1}{x+1}\right)dx = \left[\log\frac{x}{x+1}\right]_1^\infty = \log 1 - \log\frac{1}{2} = \log 2$

(6)

$\int_{-\infty}^\infty \frac{dx}{x^2+x+1} = \int_{-\infty}^\infty \frac{1}{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}dx = \left[\frac{2}{\sqrt3}\tan^{-1}\frac{x+\frac{1}{2}}{\frac{\sqrt3}{2}}\right]_{-\infty}^\infty = \frac{2}{\sqrt3}\left(\frac{\pi}{2}+\frac{\pi}{2}\right)=\frac{2}{\sqrt3}\pi$