What is the integral of 1/((6x^2+9)^4)x ?

The integral of 1/((6x^2+9)^4)x is -1/(36(6x^2+9)^3)+C

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integral of 1/((6x^2+9)^4)x

\int \frac{1}{( 6 x ^{2} + 9 ) ^{4}} x d x

- \frac{1}{36 ( 6 x ^{2} + 9 ) ^{3}} + C

Solution steps

\int \frac{1}{( 6 x ^{2} + 9 ) ^{4}} x d x

Simplify

\frac{1}{( 6 x ^{2} + 9 ) ^{4}} x : \frac{x}{( 6 x ^{2} + 9 ) ^{4}}

= \int \frac{x}{( 6 x ^{2} + 9 ) ^{4}} d x

Apply u-substitution

: \int \frac{1}{12 u ^{4}} d u

= \int \frac{1}{12 u ^{4}} d u

Take the constant out:

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

= \frac{1}{12} \cdot \int \frac{1}{u ^{4}} d u

Apply the Power Rule

: - \frac{1}{3 u ^{3}}

= \frac{1}{12} (- \frac{1}{3 u ^{3}})

Substitute back

u = 6 x^{2} + 9

= \frac{1}{12} (- \frac{1}{3 ( 6 x ^{2} + 9 ) ^{3}})

Simplify

\frac{1}{12} (- \frac{1}{3 ( 6 x ^{2} + 9 ) ^{3}}) : - \frac{1}{36 ( 6 x ^{2} + 9 ) ^{3}}

= - \frac{1}{36 ( 6 x ^{2} + 9 ) ^{3}}

Add a constant to the solution

= - \frac{1}{36 ( 6 x ^{2} + 9 ) ^{3}} + C

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What is the integral of 1/((6x^2+9)^4)x ?
The integral of 1/((6x^2+9)^4)x is -1/(36(6x^2+9)^3)+C