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

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

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

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

3 ln x^{2} + 4 - 2 ln ∣ x ∣ + C

Solution steps

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

Expand

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

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

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

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

= 2 ln x^{2} + 4 - (2 ln ∣ x ∣ - ln x^{2} + 4)

Simplify

= 3 ln x^{2} + 4 - 2 ln ∣ x ∣

Add a constant to the solution

= 3 ln x^{2} + 4 - 2 ln ∣ x ∣ + C

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