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

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

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

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

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

Solution steps

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

Simplify

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

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

Apply u-substitution

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

= \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 = 1 + x^{2}

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

Add a constant to the solution

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

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