What is the integral of (x+4)/(x^2+8x) ?

The integral of (x+4)/(x^2+8x) is 1/2 ln|x|+1/2 ln|x+8|+C

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integral of (x+4)/(x^2+8x)

\int \frac{x + 4}{x ^{2} + 8 x} d x

\frac{1}{2} ln ∣ x ∣ + \frac{1}{2} ln ∣ x + 8 ∣ + C

Solution steps

\int \frac{x + 4}{x ^{2} + 8 x} d x

Take the partial fraction of

\frac{x + 4}{x ^{2} + 8 x} : \frac{1}{2 x} + \frac{1}{2 ( x + 8 )}

= \int \frac{1}{2 x} + \frac{1}{2 ( 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}{2 x} d x + \int \frac{1}{2 ( x + 8 )} d x

\int \frac{1}{2 x} d x = \frac{1}{2} ln ∣ x ∣

\int \frac{1}{2 ( x + 8 )} d x = \frac{1}{2} ln ∣ x + 8 ∣

= \frac{1}{2} ln ∣ x ∣ + \frac{1}{2} ln ∣ x + 8 ∣

Add a constant to the solution

= \frac{1}{2} ln ∣ x ∣ + \frac{1}{2} ln ∣ x + 8 ∣ + C

x y^{^{'}} + y = 2 x

x \to - \infty lim (x e^{x})

x \to 0 lim (\frac{3}{1 + e ^{\frac{1}{x}}})

\int \frac{x ^{2}}{9 - x ^{2}} d x

f (t) = 1 + t

What is the integral of (x+4)/(x^2+8x) ?
The integral of (x+4)/(x^2+8x) is 1/2 ln|x|+1/2 ln|x+8|+C