integral of xln(x^4-1)

解答

\int x ln (x^{4} - 1) d x

\frac{1}{2} x^{2} ln (x^{4} - 1) - x^{2} + \frac{1}{2} ln x^{2} + 1 - \frac{1}{2} ln x^{2} - 1 + C

求解步骤

\int x ln (x^{4} - 1) d x

使用分布积分法

= \frac{1}{2} x^{2} ln (x^{4} - 1) - \int \frac{2 x ^{5}}{x ^{4} - 1} d x

\int \frac{2 x ^{5}}{x ^{4} - 1} d x = x^{2} - \frac{1}{2} ln x^{2} + 1 + \frac{1}{2} ln x^{2} - 1

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

化简

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

解答补常数

= \frac{1}{2} x^{2} ln (x^{4} - 1) - x^{2} + \frac{1}{2} ln x^{2} + 1 - \frac{1}{2} ln x^{2} - 1 + C

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