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

\int \frac{x - 4}{( x + 2 ) ^{2} + 16} d x

x \to 0 lim (\frac{e ^{x^{3}} - 1}{x})

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

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

f (x) = (ln (x))^{4}, at x = 7

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