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))

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

\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

= \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

\frac{\partial}{\partial x} (6 x^{4 y})

\int x^{3} 1 - x d x

\frac{d}{d x} (4 - \frac{25}{6 x ^{\frac{1}{6}}})

\frac{d}{d x} (ln (3 x^{2} + 7))

f (x) = x - \frac{1}{x}, at x = 1

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