문제 1-3의 풀이

x를 tany로 치환하여 부정적분을 계산하자 $$\int_{-1}^{1} {\frac{dx}{x^2+1}}=\int_{-1}^{1} {\frac{dtany}{sec^2y}}=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} {dy}=\frac{\pi}{2}$$--왼손은거들뿐 (talk) 09:22, May 23, 2015 (UTC)

$$\left( \frac{1}{2} arccos \frac{1-x^2}{1+x^2} \right)' = \frac{1}{x^2+1}$$이므로

$$\int_{-1}^{1} \frac{1}{x^2 +1 } dx = \left[ \frac{1}{2} arccos \frac{1-x^2}{1+x^2} \right]_{-1}^{1} = \frac{1}{2} (arccos0 - arccos0)=0$$ 따라서 구하는 값은 0이다. --Xlot (talk) 07:32, May 31, 2015 (UTC)