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

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

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

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

2 ln ∣ x ∣ - \frac{3}{2} ln x^{2} + 4 + C

Solution steps

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

Take the partial fraction of

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

= \int \frac{2}{x} - \frac{3 x}{x ^{2} + 4} d x

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

\int \frac{2}{x} d x = 2 ln ∣ x ∣

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

= 2 ln ∣ x ∣ - \frac{3}{2} ln x^{2} + 4

Add a constant to the solution

= 2 ln ∣ x ∣ - \frac{3}{2} ln x^{2} + 4 + C

\int (2 x + 4 y) d x

\int (2 e^{x} + sin (x) - cos (x)) d x

\int \frac{1}{x ( x ^{6} + 8 )} d x

\int \frac{x}{2 + 4 x ^{2}} d x

\int \frac{r}{r ^{2} - x ^{2}} d x

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