What is the integral of (-2x)/(x^2+4) ?

The integral of (-2x)/(x^2+4) is -ln|x^2+4|+C

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integral of (-2x)/(x^2+4)

\int \frac{- 2 x}{x ^{2} + 4} d x

- ln x^{2} + 4 + C

Solution steps

\int \frac{- 2 x}{x ^{2} + 4} 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}{x ^{2} + 4} 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 = x^{2} + 4

= - 2 \cdot \frac{1}{2} ln x^{2} + 4

Simplify

- 2 \cdot \frac{1}{2} ln x^{2} + 4 : - ln x^{2} + 4

= - ln x^{2} + 4

Add a constant to the solution

= - ln x^{2} + 4 + C

x \to 3 lim (x 18 - x^{2})

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

y = 2 e^{x} cos (x)

\frac{( \frac{1}{2} )}{s}

\int 3 x^{\frac{1}{8}} d x

What is the integral of (-2x)/(x^2+4) ?
The integral of (-2x)/(x^2+4) is -ln|x^2+4|+C