What is the integral of (x^3)/((16-x^4)^2) ?

The integral of (x^3)/((16-x^4)^2) is 1/(4(16-x^4))+C

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integral of (x^3)/((16-x^4)^2)

\int \frac{x ^{3}}{( 16 - x ^{4} ) ^{2}} d x

\frac{1}{4 ( 16 - x ^{4} )} + C

Solution steps

\int \frac{x ^{3}}{( 16 - x ^{4} ) ^{2}} d x

Apply u-substitution

: \int - \frac{1}{4 u ^{2}} d u

= \int - \frac{1}{4 u ^{2}} d u

Take the constant out:

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

= - \frac{1}{4} \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= - \frac{1}{4} (- \frac{1}{u})

Substitute back

u = 16 - x^{4}

= - \frac{1}{4} (- \frac{1}{16 - x ^{4}})

Simplify

- \frac{1}{4} (- \frac{1}{16 - x ^{4}}) : \frac{1}{4 ( 16 - x ^{4} )}

= \frac{1}{4 ( 16 - x ^{4} )}

Add a constant to the solution

= \frac{1}{4 ( 16 - x ^{4} )} + C

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What is the integral of (x^3)/((16-x^4)^2) ?
The integral of (x^3)/((16-x^4)^2) is 1/(4(16-x^4))+C