What is the integral of 1/(x^2+4x) ?

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

integral of 1/(x^2+4x)

\int \frac{1}{x ^{2} + 4 x} d x

\frac{1}{4} ln ∣ x ∣ - \frac{1}{4} ln ∣ x + 4 ∣ + C

Solution steps

\int \frac{1}{x ^{2} + 4 x} d x

Take the partial fraction of

\frac{1}{x ^{2} + 4 x} : \frac{1}{4 x} - \frac{1}{4 ( x + 4 )}

= \int \frac{1}{4 x} - \frac{1}{4 ( x + 4 )} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1}{4 x} d x - \int \frac{1}{4 ( x + 4 )} d x

\int \frac{1}{4 x} d x = \frac{1}{4} ln ∣ x ∣

\int \frac{1}{4 ( x + 4 )} d x = \frac{1}{4} ln ∣ x + 4 ∣

= \frac{1}{4} ln ∣ x ∣ - \frac{1}{4} ln ∣ x + 4 ∣

Add a constant to the solution

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

d er i v a t i v e \frac{3 x}{x - 9}

t an g e n t (x^{3} - 4 x)^{10}

\int \frac{1}{x ^{2} + 7 x + 12} d x

\int (sec (x) tan (x) - (sec (x))^{2}) d x

\frac{d}{d x} (\frac{[ 3 x ^{2} ]}{[ x ^{2} - 4 ]})

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