What is the integral of (x^3(1+x^2))^{-1} ?

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

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integral of (x^3(1+x^2))^{-1}

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

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

Solution steps

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

Apply exponent rule:

a^{- 1} = \frac{1}{a}

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

Take the partial fraction of

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

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

Add a constant to the solution

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

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