What is the integral of (3x^2)/((8-3x^3)^5) ?

The integral of (3x^2)/((8-3x^3)^5) is 1/(12(-3x^3+8)^4)+C

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integral of (3x^2)/((8-3x^3)^5)

\int \frac{3 x ^{2}}{( 8 - 3 x ^{3} ) ^{5}} d x

\frac{1}{12 ( - 3 x ^{3} + 8 ) ^{4}} + C

Solution steps

\int \frac{3 x ^{2}}{( 8 - 3 x ^{3} ) ^{5}} d x

Take the constant out:

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

= 3 \cdot \int \frac{x ^{2}}{( 8 - 3 x ^{3} ) ^{5}} d x

Apply u-substitution

: \int - \frac{1}{9 u ^{5}} d u

= 3 \cdot \int - \frac{1}{9 u ^{5}} d u

Take the constant out:

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

= 3 (- \frac{1}{9} \cdot \int \frac{1}{u ^{5}} d u)

Apply the Power Rule

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

= 3 (- \frac{1}{9} (- \frac{1}{4 u ^{4}}))

Substitute back

u = 8 - 3 x^{3}

= 3 (- \frac{1}{9} (- \frac{1}{4 ( 8 - 3 x ^{3} ) ^{4}}))

Simplify

3 (- \frac{1}{9} (- \frac{1}{4 ( 8 - 3 x ^{3} ) ^{4}})) : \frac{1}{12 ( - 3 x ^{3} + 8 ) ^{4}}

= \frac{1}{12 ( - 3 x ^{3} + 8 ) ^{4}}

Add a constant to the solution

= \frac{1}{12 ( - 3 x ^{3} + 8 ) ^{4}} + C

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What is the integral of (3x^2)/((8-3x^3)^5) ?
The integral of (3x^2)/((8-3x^3)^5) is 1/(12(-3x^3+8)^4)+C