What is the integral of 1/(x^2+8x) ?

The integral of 1/(x^2+8x) is 1/8 ln|x|-1/8 ln|x+8|+C

integral of 1/(x^2+8x)

\int \frac{1}{x ^{2} + 8 x} d x

\frac{1}{8} ln ∣ x ∣ - \frac{1}{8} ln ∣ x + 8 ∣ + C

Solution steps

\int \frac{1}{x ^{2} + 8 x} d x

Take the partial fraction of

\frac{1}{x ^{2} + 8 x} : \frac{1}{8 x} - \frac{1}{8 ( x + 8 )}

= \int \frac{1}{8 x} - \frac{1}{8 ( x + 8 )} 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}{8 x} d x - \int \frac{1}{8 ( x + 8 )} d x

\int \frac{1}{8 x} d x = \frac{1}{8} ln ∣ x ∣

\int \frac{1}{8 ( x + 8 )} d x = \frac{1}{8} ln ∣ x + 8 ∣

= \frac{1}{8} ln ∣ x ∣ - \frac{1}{8} ln ∣ x + 8 ∣

Add a constant to the solution

= \frac{1}{8} ln ∣ x ∣ - \frac{1}{8} ln ∣ x + 8 ∣ + C

x \to 1 lim (\frac{2 x + 1}{x + 2})

\frac{d}{d x} ((9 x^{2} - 3 x - 10) y^{^{''}} + (x^{2} + x) y^{^{'}} + (2 x - 1) y)

\frac{\partial}{\partial x} (tan (6 + 2 x^{2} y^{3} z^{2}))

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

\frac{\partial}{\partial x} (x^{2} - y^{2} + 5 x y)

What is the integral of 1/(x^2+8x) ?
The integral of 1/(x^2+8x) is 1/8 ln|x|-1/8 ln|x+8|+C