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

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

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

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

- ln 1 + x^{2} + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{1}{2 u} d u

= - 2 \cdot \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

= - 2 \cdot \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - 2 \cdot \frac{1}{2} ln ∣ u ∣

Substitute back

u = 1 + x^{2}

= - 2 \cdot \frac{1}{2} ln 1 + x^{2}

Simplify

- 2 \cdot \frac{1}{2} ln 1 + x^{2} : - ln 1 + x^{2}

= - ln 1 + x^{2}

Add a constant to the solution

= - ln 1 + x^{2} + C

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