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

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

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

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

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

Solution steps

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

Apply u-substitution

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

Substitute back

u = x^{4} + 16

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

Simplify

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

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

Add a constant to the solution

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

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