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

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

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

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

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

Solution steps

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

Take the constant out:

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

= - \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

\int \frac{1}{3 x 9 x ^{2} - 12} d x

e^{- 1}

\int_{0}^{2} e^{- x} + e d x

\int t e^{- 6 t} d t

\frac{d}{d x} (\frac{2 x ^{2} + 6 x + 6}{x})

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