What is the integral of (8x^2)/((1-x^3)^3) ?

The integral of (8x^2)/((1-x^3)^3) is 4/(3(-x^3+1)^2)+C

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integral of (8x^2)/((1-x^3)^3)

\int \frac{8 x ^{2}}{( 1 - x ^{3} ) ^{3}} d x

\frac{4}{3 ( - x ^{3} + 1 ) ^{2}} + C

Solution steps

\int \frac{8 x ^{2}}{( 1 - x ^{3} ) ^{3}} d x

Take the constant out:

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

= 8 \cdot \int \frac{x ^{2}}{( 1 - x ^{3} ) ^{3}} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Apply u-substitution

: \int - \frac{1}{v ^{3}} d v

= 8 \cdot \frac{1}{3} \cdot \int - \frac{1}{v ^{3}} d v

Take the constant out:

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

= 8 \cdot \frac{1}{3} (- \int \frac{1}{v ^{3}} d v)

Apply the Power Rule

: - \frac{1}{2 v ^{2}}

= 8 \cdot \frac{1}{3} (- (- \frac{1}{2 v ^{2}}))

Substitute back

= 8 \cdot \frac{1}{3} (- (- \frac{1}{2 ( 1 - x ^{3} ) ^{2}}))

Simplify

8 \cdot \frac{1}{3} (- (- \frac{1}{2 ( 1 - x ^{3} ) ^{2}})) : \frac{4}{3 ( - x ^{3} + 1 ) ^{2}}

= \frac{4}{3 ( - x ^{3} + 1 ) ^{2}}

Add a constant to the solution

= \frac{4}{3 ( - x ^{3} + 1 ) ^{2}} + C

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What is the integral of (8x^2)/((1-x^3)^3) ?
The integral of (8x^2)/((1-x^3)^3) is 4/(3(-x^3+1)^2)+C