What is the integral of xln(x^4-1) ?

The integral of xln(x^4-1) is 1/2 x^2ln(x^4-1)-x^2+1/2 ln|x^2+1|-1/2 ln|x^2-1|+C

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integral of xln(x^4-1)

\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

Solution steps

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

Apply Integration By Parts

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

= \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)

Simplify

= \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

Add a constant to the solution

= \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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What is the integral of xln(x^4-1) ?
The integral of xln(x^4-1) is 1/2 x^2ln(x^4-1)-x^2+1/2 ln|x^2+1|-1/2 ln|x^2-1|+C