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

The integral of (-x)/(x^2+4) is -1/2 ln|x^2+4|+C

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

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

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

Solution steps

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

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

Add a constant to the solution

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

x \to 1 lim (2 x - 1)

f (x) = 3 x - 2 x^{2}

\frac{d y}{d x} - y = e^{x}

n = 0 \sum \infty 9 (\frac{2}{3})^{n}

\frac{d y}{d t} = 9 y - y^{2} - \frac{81}{4}

What is the integral of (-x)/(x^2+4) ?
The integral of (-x)/(x^2+4) is -1/2 ln|x^2+4|+C