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

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

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integral of 1/(x^3+x)

\int \frac{1}{x ^{3} + x} d x

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

Solution steps

\int \frac{1}{x ^{3} + x} d x

Take the partial fraction of

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

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

= ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 1

Add a constant to the solution

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

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

\int (x + 5) cos (x) d x

\int x cos^{2} (9 x) d x

\frac{d y}{d x} = 4 x e^{y}

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

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