What is the integral of x/(x^3+4x) ?

The integral of x/(x^3+4x) is 1/2 arctan(x/2)+C

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integral of x/(x^3+4x)

\int \frac{x}{x ^{3} + 4 x} d x

\frac{1}{2} arctan (\frac{x}{2}) + C

Solution steps

\int \frac{x}{x ^{3} + 4 x} d x

Factor

\frac{x}{x ^{3} + 4 x}

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

Apply Integral Substitution

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

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

= \frac{1}{2} arctan (u)

Substitute back

u = \frac{x}{2}

= \frac{1}{2} arctan (\frac{x}{2})

Add a constant to the solution

= \frac{1}{2} arctan (\frac{x}{2}) + C

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