What is the integral of 1/(x(x^2+1)^3) ?

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

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integral of 1/(x(x^2+1)^3)

\int \frac{1}{x ( x ^{2} + 1 ) ^{3}} d x

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

Solution steps

\int \frac{1}{x ( x ^{2} + 1 ) ^{3}} d x

Take the partial fraction of

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

= \int \frac{1}{x} - \frac{x}{x ^{2} + 1} - \frac{x}{( x ^{2} + 1 ) ^{2}} - \frac{x}{( x ^{2} + 1 ) ^{3}} 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}{x} d x - \int \frac{x}{x ^{2} + 1} d x - \int \frac{x}{( x ^{2} + 1 ) ^{2}} d x - \int \frac{x}{( x ^{2} + 1 ) ^{3}} d x

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{x}{x ^{2} + 1} d x = \frac{1}{2} ln x^{2} + 1

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

\int \frac{x}{( x ^{2} + 1 ) ^{3}} d x = - \frac{1}{4 ( x ^{2} + 1 ) ^{2}}

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

Simplify

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

Add a constant to the solution

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

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