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

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

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

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

ln ∣ x + 4 ∣ + C

Solution steps

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

Simplify

\frac{x}{x ^{2} + 4 x} : \frac{1}{x + 4}

= \int \frac{1}{x + 4} d x

Apply u-substitution

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

= \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= ln ∣ u ∣

Substitute back

u = x + 4

= ln ∣ x + 4 ∣

Add a constant to the solution

= ln ∣ x + 4 ∣ + C

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