1/(x²+1)²의 정적분

$x = \tan \theta$ $\displaystyle \left( - \frac{\pi}{2} < \theta < \frac{\pi}{2} \right)$로 놓으면 $dx = \sec ^{2} \theta d \theta$이고 $x = 0$일 때 $\theta = 0$, $x = 1$일 때 $\displaystyle \theta = \frac{\pi}{4}$ $( x ^{2} + 1 ) ^{2} = ( \tan ^{2} \theta + 1 ) ^{2} = \sec ^{4} \theta$이므로 $\displaystyle \int _{0} ^{1} \frac{1}{( x ^{2} + 1 ) ^{2}} dx = \int _{0} ^{\frac{\pi}{4}} \frac{\sec ^{2} \theta}{\sec ^{4} \theta} d \theta = \int _{0} ^{\frac{\pi}{4}} \cos ^{2} \theta d \theta$ $\displaystyle = \int _{0} ^{\frac{\pi}{4}} \frac{1 + \cos 2 \theta}{2} d \theta = \left[ \frac{\theta}{2} + \frac{1}{4} \sin 2 \theta \right] _{0} ^{\frac{\pi}{4}} = \frac{\pi}{8} + \frac{1}{4}$ 따라서 구하는 값은 $\displaystyle \frac{\pi}{8} + \frac{1}{4}$ ①