integral of xarctan(x+1)

解答

\int x arctan (x + 1) d x

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

求解步骤

\int x arctan (x + 1) d x

使用分布积分法

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

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

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

解答补常数

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

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