What is the integral of (x^5)/((9-x^6)^2) ?

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

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integral of (x^5)/((9-x^6)^2)

\int \frac{x ^{5}}{( 9 - x ^{6} ) ^{2}} d x

\frac{1}{6 ( 9 - x ^{6} )} + C

Solution steps

\int \frac{x ^{5}}{( 9 - x ^{6} ) ^{2}} d x

Apply u-substitution

= \int - \frac{1}{6 u ^{2}} d u

Take the constant out:

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

= - \frac{1}{6} \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

= - \frac{1}{6} (- \frac{1}{u})

Substitute back

u = 9 - x^{6}

= - \frac{1}{6} (- \frac{1}{9 - x ^{6}})

Simplify

- \frac{1}{6} (- \frac{1}{9 - x ^{6}}) : \frac{1}{6 ( 9 - x ^{6} )}

= \frac{1}{6 ( 9 - x ^{6} )}

Add a constant to the solution

= \frac{1}{6 ( 9 - x ^{6} )} + C

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