What is the integral of x/((16-x^2)^2) ?

The integral of x/((16-x^2)^2) is 1/(2(16-x^2))+C

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integral of x/((16-x^2)^2)

\int \frac{x}{( 16 - x ^{2} ) ^{2}} d x

\frac{1}{2 ( 16 - x ^{2} )} + C

Solution steps

\int \frac{x}{( 16 - x ^{2} ) ^{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 = 16 - x^{2}

= - \frac{1}{2} (- \frac{1}{16 - x ^{2}})

Simplify

- \frac{1}{2} (- \frac{1}{16 - x ^{2}}) : \frac{1}{2 ( 16 - x ^{2} )}

= \frac{1}{2 ( 16 - x ^{2} )}

Add a constant to the solution

= \frac{1}{2 ( 16 - x ^{2} )} + C

What is the integral of x/((16-x^2)^2) ?
The integral of x/((16-x^2)^2) is 1/(2(16-x^2))+C