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

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

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integral of 2/(x^3+x)

\int \frac{2}{x ^{3} + x} d x

2 ln ∣ x ∣ - ln x^{2} + 1 + C

Solution steps

\int \frac{2}{x ^{3} + x} d x

Take the constant out:

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

= 2 \cdot \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}

= 2 \cdot \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

= 2 (\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

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

Simplify

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

= 2 ln ∣ x ∣ - ln x^{2} + 1

Add a constant to the solution

= 2 ln ∣ x ∣ - ln x^{2} + 1 + C

\int_{0}^{2} 4 - x^{2} d x

\frac{\partial}{\partial y} (x y^{2})

3 ln (2 x)

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

\frac{d}{d x} (1 0^{6} \cdot x^{2} \cdot e^{- x})

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