What is the integral of 1/(x^2+1)x ?

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

integral of 1/(x^2+1)x

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

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

Solution steps

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

Simplify

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

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

Apply u-substitution

= \int \frac{1}{2 u} 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} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

u = x^{2} + 1

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

Add a constant to the solution

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

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