What is the integral of (x^2)/((2-2x^3)^3) ?

The integral of (x^2)/((2-2x^3)^3) is 1/(48(x-1)^2(x^2+x+1)^2)+C

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integral of (x^2)/((2-2x^3)^3)

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

\frac{1}{48 ( x - 1 ) ^{2} ( x ^{2} + x + 1 ) ^{2}} + C

Solution steps

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

Apply u-substitution

= \int \frac{1}{3 ( 2 - 2 u ) ^{3}} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{1}{( 2 - 2 u ) ^{3}} d u

Apply Integration By Parts

= \frac{1}{3} (\frac{u}{( 2 - 2 u ) ^{3}} - \int \frac{3 u}{8 ( u - 1 ) ^{4}} d u)

\int \frac{3 u}{8 ( u - 1 ) ^{4}} d u = \frac{- 3 u + 1}{16 ( u - 1 ) ^{3}}

= \frac{1}{3} (\frac{u}{( 2 - 2 u ) ^{3}} - \frac{- 3 u + 1}{16 ( u - 1 ) ^{3}})

Substitute back

u = x^{3}

= \frac{1}{3} (\frac{x ^{3}}{( 2 - 2 x ^{3} ) ^{3}} - \frac{- 3 x ^{3} + 1}{16 ( x ^{3} - 1 ) ^{3}})

Simplify

\frac{1}{3} (\frac{x ^{3}}{( 2 - 2 x ^{3} ) ^{3}} - \frac{- 3 x ^{3} + 1}{16 ( x ^{3} - 1 ) ^{3}}) : \frac{1}{48 ( x - 1 ) ^{2} ( x ^{2} + x + 1 ) ^{2}}

= \frac{1}{48 ( x - 1 ) ^{2} ( x ^{2} + x + 1 ) ^{2}}

Add a constant to the solution

= \frac{1}{48 ( x - 1 ) ^{2} ( x ^{2} + x + 1 ) ^{2}} + C

What is the integral of (x^2)/((2-2x^3)^3) ?
The integral of (x^2)/((2-2x^3)^3) is 1/(48(x-1)^2(x^2+x+1)^2)+C