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integral of (x*arctan(x))/((1+x^2)^2)

解答

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

解答

\frac{1}{2} (- \frac{1}{2} arctan (x) cos (2 arctan (x)) + \frac{1}{4} sin (2 arctan (x))) + C

求解步骤

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

使用换元积分法

= \int \frac{u tan ( u )}{sec ^{2} ( u )} d u

使用三角恒等式改写

= \int \frac{1}{2} u sin (2 u) d u

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int u sin (2 u) d u

使用分布积分法

= \frac{1}{2} (- \frac{1}{2} u cos (2 u) - \int - \frac{1}{2} cos (2 u) d u)

\int - \frac{1}{2} cos (2 u) d u = - \frac{1}{4} sin (2 u)

= \frac{1}{2} (- \frac{1}{2} u cos (2 u) - (- \frac{1}{4} sin (2 u)))

u = arctan (x)

代回

= \frac{1}{2} (- \frac{1}{2} arctan (x) cos (2 arctan (x)) - (- \frac{1}{4} sin (2 arctan (x))))

化简

= \frac{1}{2} (- \frac{1}{2} arctan (x) cos (2 arctan (x)) + \frac{1}{4} sin (2 arctan (x)))

解答补常数

= \frac{1}{2} (- \frac{1}{2} arctan (x) cos (2 arctan (x)) + \frac{1}{4} sin (2 arctan (x))) + C

作图

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