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

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

\int \frac{1}{( x - 2 ) ^{2} ( x ^{2} - 4 x + 3 )} d x

f (x) = x^{3} + 4 x^{2} + 3 x

\int \frac{9 - x ^{2}}{x} d x

\frac{d}{d x} ((x - 1) (x^{2} + 2)^{3})

\frac{\partial}{\partial x} (ln (2 + x^{2} y^{2}))

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