第5章 5.12

(1)

$\int_0^1 \frac{dx}{x^2} = \left[-\frac{1}{x}\right]_0^1 = -1+\infty \qquad (\infty \text{は} x \to +0 \text{のとき})$

よって発散する。

(2)

$\int_1^\infty \frac{dx}{x^2} = \left[-\frac{1}{x}\right]_1^\infty = 0+1=1$

(3)

$\int_0^1 \frac{dx}{\sqrt{1-x}} = \left[-2\sqrt{1-x}\right]_0^1 = 0+2=2$

(4)

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

(5)

$\int_0^\infty \frac{dx}{9x^2+4} = \int_0^\infty \frac{\dfrac{1}{9}}{x^2+\dfrac{4}{9}}dx = \frac{1}{9}\left[\frac{1}{\frac{2}{3}}\tan^{-1}\frac{x}{\frac{2}{3}}\right]_0^\infty = \frac{1}{6}\cdot\frac{\pi}{2}=\frac{\pi}{12}$

(6)

$\int_1^\infty \frac{dx}{x(x+1)} = \int_1^\infty \left(\frac{1}{x}-\frac{1}{x+1}\right)dx = \left[\log\frac{x}{x+1}\right]_1^\infty = 0-\log\frac{1}{2}=\log 2$