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

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

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integral of (6x)/((9+x^2)^3)

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{1}{2 u ^{3}} d u

= 6 \cdot \int \frac{1}{2 u ^{3}} d u

Take the constant out:

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

= 6 \cdot \frac{1}{2} \cdot \int \frac{1}{u ^{3}} d u

Apply the Power Rule

: - \frac{1}{2 u ^{2}}

= 6 \cdot \frac{1}{2} (- \frac{1}{2 u ^{2}})

Substitute back

u = 9 + x^{2}

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

Simplify

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

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

Add a constant to the solution

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

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