What is the integral of (x^2)/((1-x^3)^5) ?

The integral of (x^2)/((1-x^3)^5) is 1/(12(1-x^3)^4)+C

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integral of (x^2)/((1-x^3)^5)

\int \frac{x ^{2}}{( 1 - x ^{3} ) ^{5}} d x

\frac{1}{12 ( 1 - x ^{3} ) ^{4}} + C

Solution steps

\int \frac{x ^{2}}{( 1 - x ^{3} ) ^{5}} d x

Apply u-substitution

: \int - \frac{1}{3 u ^{5}} d u

= \int - \frac{1}{3 u ^{5}} 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}{u ^{5}} d u

Apply the Power Rule

: - \frac{1}{4 u ^{4}}

= - \frac{1}{3} (- \frac{1}{4 u ^{4}})

Substitute back

u = 1 - x^{3}

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

Simplify

- \frac{1}{3} (- \frac{1}{4 ( 1 - x ^{3} ) ^{4}}) : \frac{1}{12 ( 1 - x ^{3} ) ^{4}}

= \frac{1}{12 ( 1 - x ^{3} ) ^{4}}

Add a constant to the solution

= \frac{1}{12 ( 1 - x ^{3} ) ^{4}} + C

\frac{d}{d x} (\frac{1}{x ^{2} + 3})

\int \frac{cos ( x )}{8 - sin ( x )} d x

\int \frac{7}{9 + 16 r ^{2}} d r

\frac{d y}{d x} = 4 y + 3 e^{8 x} (x + 1), y (0) = 0

What is the integral of (x^2)/((1-x^3)^5) ?
The integral of (x^2)/((1-x^3)^5) is 1/(12(1-x^3)^4)+C