What is the integral of (x^6)/((9-x^7)^2) ?

The integral of (x^6)/((9-x^7)^2) is 1/(7(9-x^7))+C

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integral of (x^6)/((9-x^7)^2)

\int \frac{x ^{6}}{( 9 - x ^{7} ) ^{2}} d x

\frac{1}{7 ( 9 - x ^{7} )} + C

Solution steps

\int \frac{x ^{6}}{( 9 - x ^{7} ) ^{2}} d x

Apply u-substitution

= \int - \frac{1}{7 u ^{2}} d u

Take the constant out:

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

= - \frac{1}{7} \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

= - \frac{1}{7} (- \frac{1}{u})

Substitute back

u = 9 - x^{7}

= - \frac{1}{7} (- \frac{1}{9 - x ^{7}})

Simplify

- \frac{1}{7} (- \frac{1}{9 - x ^{7}}) : \frac{1}{7 ( 9 - x ^{7} )}

= \frac{1}{7 ( 9 - x ^{7} )}

Add a constant to the solution

= \frac{1}{7 ( 9 - x ^{7} )} + C

What is the integral of (x^6)/((9-x^7)^2) ?
The integral of (x^6)/((9-x^7)^2) is 1/(7(9-x^7))+C