integral of rivative of 1/(sqrt(x^2+1))

解

\int \frac{d}{d x} (\frac{1}{x ^{2} + 1}) d x

x \frac{d}{d x} (\frac{1}{x ^{2} + 1}) + \frac{2}{x ^{2} + 1} - \frac{1}{( x ^{2} + 1 ) ^{\frac{3}{2}}} + C

解答ステップ

\int \frac{d}{d x} (\frac{1}{x ^{2} + 1}) d x

部分積分を適用する

= x \frac{d}{d x} (\frac{1}{x ^{2} + 1}) - \int - \frac{x ( - 2 x ^{2} + 1 )}{( x ^{2} + 1 ) ^{\frac{5}{2}}} d x

\int - \frac{x ( - 2 x ^{2} + 1 )}{( x ^{2} + 1 ) ^{\frac{5}{2}}} d x = - \frac{2}{x ^{2} + 1} + \frac{1}{( x ^{2} + 1 ) ^{\frac{3}{2}}}

= x \frac{d}{d x} (\frac{1}{x ^{2} + 1}) - (- \frac{2}{x ^{2} + 1} + \frac{1}{( x ^{2} + 1 ) ^{\frac{3}{2}}})

簡素化

= x \frac{d}{d x} (\frac{1}{x ^{2} + 1}) + \frac{2}{x ^{2} + 1} - \frac{1}{( x ^{2} + 1 ) ^{\frac{3}{2}}}

定数を解答に追加する

= x \frac{d}{d x} (\frac{1}{x ^{2} + 1}) + \frac{2}{x ^{2} + 1} - \frac{1}{( x ^{2} + 1 ) ^{\frac{3}{2}}} + C