What is the integral of 1/(x^2+6x) ?

The integral of 1/(x^2+6x) is 1/6 ln|x|-1/6 ln|x+6|+C

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integral of 1/(x^2+6x)

\int \frac{1}{x ^{2} + 6 x} d x

\frac{1}{6} ln ∣ x ∣ - \frac{1}{6} ln ∣ x + 6 ∣ + C

Solution steps

\int \frac{1}{x ^{2} + 6 x} d x

Take the partial fraction of

\frac{1}{x ^{2} + 6 x} : \frac{1}{6 x} - \frac{1}{6 ( x + 6 )}

= \int \frac{1}{6 x} - \frac{1}{6 ( x + 6 )} 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}{6 x} d x - \int \frac{1}{6 ( x + 6 )} d x

\int \frac{1}{6 x} d x = \frac{1}{6} ln ∣ x ∣

\int \frac{1}{6 ( x + 6 )} d x = \frac{1}{6} ln ∣ x + 6 ∣

= \frac{1}{6} ln ∣ x ∣ - \frac{1}{6} ln ∣ x + 6 ∣

Add a constant to the solution

= \frac{1}{6} ln ∣ x ∣ - \frac{1}{6} ln ∣ x + 6 ∣ + C

\int (x + 1) d x

\frac{d}{d z} (9 z^{2} e^{z})

(2, - 3), (4, 4)

\frac{\partial}{\partial x} (z y^{3})

\frac{d}{d x} (- 4 \cdot 3^{x})

What is the integral of 1/(x^2+6x) ?
The integral of 1/(x^2+6x) is 1/6 ln|x|-1/6 ln|x+6|+C