What is the integral of (8-5x)/(x^2-12x-32) ?

The integral of (8-5x)/(x^2-12x-32) is

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integral of (8-5x)/(x^2-12x-32)

\int \frac{8 - 5 x}{x ^{2} - 12 x - 32} d x

- \frac{2}{17} (ln \frac{x - 6}{2 17} + 1 - ln \frac{x - 6}{2 17} - 1) - 5 (\frac{1}{2} ln x^{2} - 12 x - 32 - \frac{3}{2 17} (ln \frac{1}{2 17} (x - 6) + 1 - ln \frac{1}{2 17} (x - 6) - 1)) + C

Solution steps

\int \frac{8 - 5 x}{x ^{2} - 12 x - 32} d x

Expand

\frac{8 - 5 x}{x ^{2} - 12 x - 32} : \frac{8}{x ^{2} - 12 x - 32} - \frac{5 x}{x ^{2} - 12 x - 32}

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{8}{x ^{2} - 12 x - 32} d x - \int \frac{5 x}{x ^{2} - 12 x - 32} d x

\int \frac{8}{x ^{2} - 12 x - 32} d x = - \frac{2}{17} (ln \frac{x - 6}{2 17} + 1 - ln \frac{x - 6}{2 17} - 1)

\int \frac{5 x}{x ^{2} - 12 x - 32} d x = 5 (\frac{1}{2} ln x^{2} - 12 x - 32 - \frac{3}{2 17} (ln \frac{1}{2 17} (x - 6) + 1 - ln \frac{1}{2 17} (x - 6) - 1))

= - \frac{2}{17} (ln \frac{x - 6}{2 17} + 1 - ln \frac{x - 6}{2 17} - 1) - 5 (\frac{1}{2} ln x^{2} - 12 x - 32 - \frac{3}{2 17} (ln \frac{1}{2 17} (x - 6) + 1 - ln \frac{1}{2 17} (x - 6) - 1))

Add a constant to the solution

= - \frac{2}{17} (ln \frac{x - 6}{2 17} + 1 - ln \frac{x - 6}{2 17} - 1) - 5 (\frac{1}{2} ln x^{2} - 12 x - 32 - \frac{3}{2 17} (ln \frac{1}{2 17} (x - 6) + 1 - ln \frac{1}{2 17} (x - 6) - 1)) + C

\int \frac{t}{( t ^{2} + 1 ) ^{\frac{13}{2}}} d t

x \to 0 lim (\frac{tan ( 2 x )}{x})

\frac{\partial}{\partial x} (cos (4 y - 2 x))

x^{^{'}} = 2 - x, x (5) = 4

x \to 0 lim ((x^{4}) cos (\frac{2}{x}))

What is the integral of (8-5x)/(x^2-12x-32) ?
The integral of (8-5x)/(x^2-12x-32) is