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受欢迎的微积分 >

integral of w^2ln(w^6-1)

解答

\int w^{2} ln (w^{6} - 1) d w

解答

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

求解步骤

\int w^{2} ln (w^{6} - 1) d w

使用换元积分法

= \int \frac{ln ( u - 1 )}{6 u} d u

提出常数:

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

= \frac{1}{6} \cdot \int \frac{ln ( u - 1 )}{u} d u

使用分布积分法

= \frac{1}{6} (2 u ln (u - 1) - \int \frac{2 u}{u - 1} d u)

\int \frac{2 u}{u - 1} d u = - 4 (\frac{1}{2} ln u + 1 - \frac{1}{2} ln u - 1 - u)

= \frac{1}{6} (2 u ln (u - 1) - (- 4 (\frac{1}{2} ln u + 1 - \frac{1}{2} ln u - 1 - u)))

u = w^{6}

代回

= \frac{1}{6} (2 w^{6} ln (w^{6} - 1) - (- 4 (\frac{1}{2} ln w^{6} + 1 - \frac{1}{2} ln w^{6} - 1 - w^{6})))

化简

\frac{1}{6} (2 w^{6} ln (w^{6} - 1) - (- 4 (\frac{1}{2} ln w^{6} + 1 - \frac{1}{2} ln w^{6} - 1 - w^{6}))) : \frac{1}{3} (w^{3} ln (w^{6} - 1) - 2 w^{3} + ln w^{3} + 1 - ln w^{3} - 1)

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

解答补常数

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

作图

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