What is the integral of 1/(x(x^2+16)) ?

The integral of 1/(x(x^2+16)) is 1/16 ln|x|-1/32 ln|x^2+16|+C

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integral of 1/(x(x^2+16))

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

\frac{1}{16} ln ∣ x ∣ - \frac{1}{32} ln x^{2} + 16 + C

Solution steps

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

Take the partial fraction of

\frac{1}{x ( x ^{2} + 16 )} : \frac{1}{16 x} - \frac{x}{16 ( x ^{2} + 16 )}

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

\int \frac{1}{16 x} d x = \frac{1}{16} ln ∣ x ∣

\int \frac{x}{16 ( x ^{2} + 16 )} d x = \frac{1}{32} ln x^{2} + 16

= \frac{1}{16} ln ∣ x ∣ - \frac{1}{32} ln x^{2} + 16

Add a constant to the solution

= \frac{1}{16} ln ∣ x ∣ - \frac{1}{32} ln x^{2} + 16 + C

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What is the integral of 1/(x(x^2+16)) ?
The integral of 1/(x(x^2+16)) is 1/16 ln|x|-1/32 ln|x^2+16|+C