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

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{0}

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x \to 4 lim (8 - 3 x)

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