What is the integral of (9x^8+6)/((x^9+6x)^2) ?

The integral of (9x^8+6)/((x^9+6x)^2) is -1/(x^9+6x)+C

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integral of (9x^8+6)/((x^9+6x)^2)

\int \frac{9 x ^{8} + 6}{( x ^{9} + 6 x ) ^{2}} d x

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

Solution steps

\int \frac{9 x ^{8} + 6}{( x ^{9} + 6 x ) ^{2}} d x

Apply u-substitution

: \int \frac{1}{u ^{2}} d u

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

Apply the Power Rule

: - \frac{1}{u}

= - \frac{1}{u}

Substitute back

u = x^{9} + 6 x

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

Add a constant to the solution

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

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What is the integral of (9x^8+6)/((x^9+6x)^2) ?
The integral of (9x^8+6)/((x^9+6x)^2) is -1/(x^9+6x)+C