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 + x e^{x})

\int \frac{x ^{2} + 6 x}{x + 3} d x

\int (0.1 x + 1) e^{0.03 x} d x

f (x) = (x - 6) (3 x + 5)

\frac{d}{d x} (e^{2 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