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

The integral of (16)/(x^3+4x) is 16(1/4 ln|x|-1/8 ln|x^2+4|)+C

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

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

16 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln x^{2} + 4) + C

Solution steps

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

Take the constant out:

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

= 16 \cdot \int \frac{1}{x ^{3} + 4 x} d x

Take the partial fraction of

\frac{1}{x ^{3} + 4 x} : \frac{1}{4 x} - \frac{x}{4 ( x ^{2} + 4 )}

= 16 \cdot \int \frac{1}{4 x} - \frac{x}{4 ( 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

= 16 (\int \frac{1}{4 x} d x - \int \frac{x}{4 ( x ^{2} + 4 )} d x)

\int \frac{1}{4 x} d x = \frac{1}{4} ln ∣ x ∣

\int \frac{x}{4 ( x ^{2} + 4 )} d x = \frac{1}{8} ln x^{2} + 4

= 16 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln x^{2} + 4)

Add a constant to the solution

= 16 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln x^{2} + 4) + C

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What is the integral of (16)/(x^3+4x) ?
The integral of (16)/(x^3+4x) is 16(1/4 ln|x|-1/8 ln|x^2+4|)+C