What is the integral of (15x^2)/((9-2x^3)^6) ?

The integral of (15x^2)/((9-2x^3)^6) is 1/(2(-2x^3+9)^5)+C

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integral of (15x^2)/((9-2x^3)^6)

\int \frac{15 x ^{2}}{( 9 - 2 x ^{3} ) ^{6}} d x

\frac{1}{2 ( - 2 x ^{3} + 9 ) ^{5}} + C

Solution steps

\int \frac{15 x ^{2}}{( 9 - 2 x ^{3} ) ^{6}} d x

Take the constant out:

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

= 15 \cdot \int \frac{x ^{2}}{( 9 - 2 x ^{3} ) ^{6}} d x

Apply u-substitution

: \int - \frac{1}{6 u ^{6}} d u

= 15 \cdot \int - \frac{1}{6 u ^{6}} d u

Take the constant out:

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

= 15 (- \frac{1}{6} \cdot \int \frac{1}{u ^{6}} d u)

Apply the Power Rule

: - \frac{1}{5 u ^{5}}

= 15 (- \frac{1}{6} (- \frac{1}{5 u ^{5}}))

Substitute back

u = 9 - 2 x^{3}

= 15 (- \frac{1}{6} (- \frac{1}{5 ( 9 - 2 x ^{3} ) ^{5}}))

Simplify

15 (- \frac{1}{6} (- \frac{1}{5 ( 9 - 2 x ^{3} ) ^{5}})) : \frac{1}{2 ( - 2 x ^{3} + 9 ) ^{5}}

= \frac{1}{2 ( - 2 x ^{3} + 9 ) ^{5}}

Add a constant to the solution

= \frac{1}{2 ( - 2 x ^{3} + 9 ) ^{5}} + C

\int - e

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What is the integral of (15x^2)/((9-2x^3)^6) ?
The integral of (15x^2)/((9-2x^3)^6) is 1/(2(-2x^3+9)^5)+C