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

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

Popular >

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

Take the constant out:

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

= - \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{d}{d x} (x^{6} (x - 1)^{5})

\frac{1}{ln ( 1 + x )} - \frac{1}{x}

\frac{\partial}{\partial u} (2 sin (u) sin (v))

\int x^{2} sin (n x) d x

n = 0 \sum \infty (- 2)^{n}

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