integral of 1/(xsqrt(400x^2-400))

解

\int \frac{1}{x 400 x ^{2} - 400} d x

\frac{1}{20} arctan (x^{2} - 1) + C

解答ステップ

\int \frac{1}{x 400 x ^{2} - 400} d x

部分積分を適用する

= \frac{1}{20 x ^{2} - 1} - \int - \frac{2 x ^{2} - 1}{20 x ( x ^{2} - 1 ) x ^{2} - 1} d x

\int - \frac{2 x ^{2} - 1}{20 x ( x ^{2} - 1 ) x ^{2} - 1} d x = - \frac{1}{20} (- \frac{1}{x ^{2} - 1} + arctan (x^{2} - 1))

= \frac{1}{20 x ^{2} - 1} - (- \frac{1}{20} (- \frac{1}{x ^{2} - 1} + arctan (x^{2} - 1)))

簡素化

\frac{1}{20 x ^{2} - 1} - (- \frac{1}{20} (- \frac{1}{x ^{2} - 1} + arctan (x^{2} - 1))) : \frac{1}{20} arctan (x^{2} - 1)

= \frac{1}{20} arctan (x^{2} - 1)

定数を解答に追加する

= \frac{1}{20} arctan (x^{2} - 1) + C