What is the integral of (8x^3+16x)/((x^2+4)^2) ?

The integral of (8x^3+16x)/((x^2+4)^2) is 4ln|x^2+4|+8/(x^2+4)+C

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integral of (8x^3+16x)/((x^2+4)^2)

\int \frac{8 x ^{3} + 16 x}{( x ^{2} + 4 ) ^{2}} d x

4 ln x^{2} + 4 + \frac{8}{x ^{2} + 4} + C

Solution steps

\int \frac{8 x ^{3} + 16 x}{( x ^{2} + 4 ) ^{2}} d x

Expand

\frac{8 x ^{3} + 16 x}{( x ^{2} + 4 ) ^{2}} : \frac{8 x ^{3}}{( x ^{2} + 4 ) ^{2}} + \frac{16 x}{( x ^{2} + 4 ) ^{2}}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{8 x ^{3}}{( x ^{2} + 4 ) ^{2}} d x + \int \frac{16 x}{( x ^{2} + 4 ) ^{2}} d x

\int \frac{8 x ^{3}}{( x ^{2} + 4 ) ^{2}} d x = 4 (ln x^{2} + 4 + \frac{4}{x ^{2} + 4})

\int \frac{16 x}{( x ^{2} + 4 ) ^{2}} d x = - \frac{8}{x ^{2} + 4}

= 4 (ln x^{2} + 4 + \frac{4}{x ^{2} + 4}) - \frac{8}{x ^{2} + 4}

Simplify

4 (ln x^{2} + 4 + \frac{4}{x ^{2} + 4}) - \frac{8}{x ^{2} + 4} : 4 ln x^{2} + 4 + \frac{8}{x ^{2} + 4}

= 4 ln x^{2} + 4 + \frac{8}{x ^{2} + 4}

Add a constant to the solution

= 4 ln x^{2} + 4 + \frac{8}{x ^{2} + 4} + C

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What is the integral of (8x^3+16x)/((x^2+4)^2) ?
The integral of (8x^3+16x)/((x^2+4)^2) is 4ln|x^2+4|+8/(x^2+4)+C