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

y^{^{''}} + 16 y = 2 sin (4 x)

y^{^{''}} - k^{2} y = 0

x^{4} - 8 x^{3} + 2

\int sin (π x) d x

\int \frac{4 x ^{2} - 3 x - 4}{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