What is the integral of x^2ln(x^6-1) ?

The integral of x^2ln(x^6-1) is 1/3 (x^3ln(x^6-1)-2x^3+ln|x^3+1|-ln|x^3-1|)+C

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

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

\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

Apply Integration By Parts

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

= \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)))

Substitute back

u = x^{6}

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

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial y} (5 x^{4 y})

\frac{\partial}{\partial x} (\frac{( x - 5 y )}{5 x y + 7})

\frac{( 2 - \frac{1}{x} )}{x - 3}

\int \frac{27}{( 1 - x ^{2} ) ^{\frac{3}{2}}} d x

\frac{d}{d x} (cot (sin (x)))

What is the integral of x^2ln(x^6-1) ?
The integral of x^2ln(x^6-1) is 1/3 (x^3ln(x^6-1)-2x^3+ln|x^3+1|-ln|x^3-1|)+C