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

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

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

integral

\frac{x}{( x ^{2} + 1 )}

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

Solution steps

\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

in t e g r a l \frac{1}{( 1 - x ^{2} )}

in t e g r a l \frac{x ^{3}}{( x + 1 ) ^{2}}

in t e g r a l (x - 1) (x + 3)^{3}

in t e g r a l (e^{y})^{2}

in t e g r a l \frac{1}{( x ^{3} + x ^{2} + x )}

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