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

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

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

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \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 )}

= 4 \cdot \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

= 4 (\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 ∣

= 4 (\frac{1}{8} ln ∣ x ∣ - \frac{1}{8} ln ∣ x + 8 ∣)

Add a constant to the solution

= 4 (\frac{1}{8} ln ∣ x ∣ - \frac{1}{8} ln ∣ x + 8 ∣) + C

\frac{d}{d x} (x e^{x y} cos (2 x) - 3)

\int_{- 1}^{1} \frac{3 x}{( x + 6 ) ^{2}} d x

\int 4 x y d x

e^{x}

\int 3 x x d x

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