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

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

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

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

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

Solution steps

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

Apply u-substitution

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

Take the constant out:

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

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

Apply the Power Rule

= \frac{1}{2} (- \frac{1}{u})

Substitute back

u = x^{2} + 9

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

Simplify

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

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

Add a constant to the solution

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

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