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

解

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

- \frac{1}{6} x^{3} + \frac{x ^{3}}{2} - \frac{1}{2} x^{2} x^{2} - 1 + \frac{1}{6} (x^{2} - 1)^{\frac{3}{2}} + \frac{1}{2} x^{2} - 1 + C

解答ステップ

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

部分積分を適用する

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

\int \frac{1}{2} (x^{2} - x x^{2} - 1 + ln x + x^{2} - 1) d x = \frac{1}{6} (x^{3} - (x^{2} - 1)^{\frac{3}{2}} + 3 x ln x + x^{2} - 1 - 3 x^{2} - 1)

= x (\frac{x ^{2}}{2} - \frac{1}{2} x x^{2} - 1 + \frac{1}{2} ln x + x^{2} - 1) - \frac{1}{6} (x^{3} - (x^{2} - 1)^{\frac{3}{2}} + 3 x ln x + x^{2} - 1 - 3 x^{2} - 1)

簡素化

x (\frac{x ^{2}}{2} - \frac{1}{2} x x^{2} - 1 + \frac{1}{2} ln x + x^{2} - 1) - \frac{1}{6} (x^{3} - (x^{2} - 1)^{\frac{3}{2}} + 3 x ln x + x^{2} - 1 - 3 x^{2} - 1) : - \frac{1}{6} x^{3} + \frac{x ^{3}}{2} - \frac{1}{2} x^{2} x^{2} - 1 + \frac{1}{6} (x^{2} - 1)^{\frac{3}{2}} + \frac{1}{2} x^{2} - 1

= - \frac{1}{6} x^{3} + \frac{x ^{3}}{2} - \frac{1}{2} x^{2} x^{2} - 1 + \frac{1}{6} (x^{2} - 1)^{\frac{3}{2}} + \frac{1}{2} x^{2} - 1

定数を解答に追加する

= - \frac{1}{6} x^{3} + \frac{x ^{3}}{2} - \frac{1}{2} x^{2} x^{2} - 1 + \frac{1}{6} (x^{2} - 1)^{\frac{3}{2}} + \frac{1}{2} x^{2} - 1 + C