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

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

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

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

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

Solution steps

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

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 = w^{6}

= \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})))

Simplify

\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)

Add a constant to the solution

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

\int \frac{1 - y ^{2}}{3 y} d y

y^{^{''}} + y = x^{2} + 1

\int \frac{8}{x ^{4} + 4} d x

\int_{0}^{1} x e^{- x} d x

\int \frac{x ^{2} - 3 x + 7}{( x ^{2} - 4 x + 6 ) ^{2}} d x

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