第4章 4.11

(1)

$\frac{\sqrt{x}}{1+x}$

$\int \frac{\sqrt{x}}{1+x}dx \qquad x=t^2 \quad \text{とおくと} \quad dx=2tdt \quad \text{これから}$

$=\int \frac{t}{1+t^2}\cdot 2tdt=\int \frac{2t^2}{1+t^2}dt=\int 2\left(1-\frac{1}{1+t^2}\right)dt=2t-2\tan^{-1}t+C \qquad \text{元に戻して}$

$=2\sqrt{x}-2\tan^{-1}\sqrt{x}+C \qquad (C\text{は積分定数})$

(2)

$\sqrt{\frac{x-1}{2-x}}$

$\int \sqrt{\frac{x-1}{2-x}}dx \qquad \sqrt{\frac{x-1}{2-x}}=t \quad \text{とおくと} \qquad \frac{\dfrac{\sqrt{2-x}}{2\sqrt{x-1}}+\dfrac{\sqrt{x-1}}{2\sqrt{2-x}}}{2-x}dx=dt$

$\text{整理して}\ \frac{1}{2(2-x)\sqrt{(x-1)(2-x)}}dx=dt \qquad \text{ここで}\ \sqrt{\frac{x-1}{2-x}}=t \quad \text{から} \quad x=2-\frac{1}{t^2+1}$

これを代入して

$\frac{1}{2\left(2-2+\dfrac{1}{t^2+1}\right)\sqrt{\left(2-\dfrac{1}{t^2+1}-1\right)\left(2-2+\dfrac{1}{t^2+1}\right)}}dx=dt \qquad dx=\frac{2t}{(t^2+1)^2}dt \qquad \text{これから}$

$=\int t\cdot \frac{2t}{(t^2+1)^2}dt=\int \frac{2t^2}{(t^2+1)^2}dt=-\frac{t}{t^2+1}+\tan^{-1}t+C \qquad (4.13(2)\text{から})$

$=\tan^{-1}\sqrt{\frac{x-1}{2-x}}-\frac{\sqrt{\dfrac{x-1}{2-x}}}{\dfrac{x-1}{2-x}+1}+C=\tan^{-1}\sqrt{\frac{x-1}{2-x}}-\sqrt{(x-1)(2-x)}+C \qquad (C\text{は積分定数})$

(3)

$\frac{1}{\sqrt{x-x^2}}$

$\int \frac{1}{\sqrt{x-x^2}}dx = \int \frac{1}{\sqrt{\frac{1}{4}-\left(x-\frac{1}{2}\right)^2}}dx = \sin^{-1}\frac{x-\frac{1}{2}}{\frac{1}{2}}+C = \sin^{-1}(2x-1)+C \qquad \text{（C\text{は積分定数}）}$

(4)

$\frac{1}{x^2\sqrt{x^2-1}}$

$\int \frac{1}{x^2\sqrt{x^2-1}}dx \qquad x=\frac{1}{\cos t} \text{とおくと} \qquad dx = \frac{\sin t}{\cos^2 t}dt \qquad \text{これから}$

$=\int \frac{1}{\frac{1}{\cos^2 t}\sqrt{\frac{1}{\cos^2 t}-1}}\cdot \frac{\sin t}{\cos^2 t}dt = \int \frac{\sin t}{\sqrt{\frac{\sin^2 t}{\cos^2 t}}}dt = \int \cos t \, dt = \sin t + C \qquad \text{元に戻して}$

$=\sin\left(\cos^{-1}\frac{1}{x}\right)+C = \frac{\sqrt{x^2-1}}{x}+C \qquad \text{（C\text{は積分定数}）}$