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

24 x (3 x^{2} + 4)^{3}

\int 4 x + \frac{3}{x} d x

cos (2 t)

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

x^{2} y^{^{'}} - x = 1 + y + x y

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