What is the integral of (2x-1)/(x^3-16x^2-64x) ?

The integral of (2x-1)/(x^3-16x^2-64x) is

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integral of (2x-1)/(x^3-16x^2-64x)

\int \frac{2 x - 1}{x ^{3} - 16 x ^{2} - 64 x} d x

- \frac{1}{8 2} ln \frac{1}{8 2} (x - 8) + 1 + \frac{1}{8 2} ln \frac{1}{8 2} (x - 8) - 1 + \frac{1}{64} ln ∣ x ∣ - \frac{1}{128 2} (2 ln x^{2} - 16 x - 64 - ln \frac{1}{8 2} x - \frac{1}{2} + 1 + ln \frac{1}{8 2} x - \frac{1}{2} - 1 + 2 ln \frac{x}{8 2} - \frac{1}{2} + 1 - 2 ln \frac{x}{8 2} - \frac{1}{2} - 1) + C

Solution steps

\int \frac{2 x - 1}{x ^{3} - 16 x ^{2} - 64 x} d x

Expand

\frac{2 x - 1}{x ^{3} - 16 x ^{2} - 64 x} : \frac{2 x}{x ^{3} - 16 x ^{2} - 64 x} - \frac{1}{x ^{3} - 16 x ^{2} - 64 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{2 x}{x ^{3} - 16 x ^{2} - 64 x} d x - \int \frac{1}{x ^{3} - 16 x ^{2} - 64 x} d x

\int \frac{2 x}{x ^{3} - 16 x ^{2} - 64 x} d x = - \frac{1}{8 2} (ln \frac{x - 8}{8 2} + 1 - ln \frac{x - 8}{8 2} - 1)

\int \frac{1}{x ^{3} - 16 x ^{2} - 64 x} d x = - \frac{1}{64} ln ∣ x ∣ + \frac{1}{64} (\frac{1}{2} ln x^{2} - 16 x - 64 - \frac{1}{2 2} (ln \frac{1}{8 2} x - \frac{1}{2} + 1 - ln \frac{1}{8 2} x - \frac{1}{2} - 1) + 2 (\frac{1}{2} ln \frac{1}{8 2} x - \frac{1}{2} + 1 - \frac{1}{2} ln \frac{1}{8 2} x - \frac{1}{2} - 1))

= - \frac{1}{8 2} (ln \frac{x - 8}{8 2} + 1 - ln \frac{x - 8}{8 2} - 1) - (- \frac{1}{64} ln ∣ x ∣ + \frac{1}{64} (\frac{1}{2} ln x^{2} - 16 x - 64 - \frac{1}{2 2} (ln \frac{1}{8 2} x - \frac{1}{2} + 1 - ln \frac{1}{8 2} x - \frac{1}{2} - 1) + 2 (\frac{1}{2} ln \frac{1}{8 2} x - \frac{1}{2} + 1 - \frac{1}{2} ln \frac{1}{8 2} x - \frac{1}{2} - 1)))

Simplify

- \frac{1}{8 2} (ln \frac{x - 8}{8 2} + 1 - ln \frac{x - 8}{8 2} - 1) - (- \frac{1}{64} ln ∣ x ∣ + \frac{1}{64} (\frac{1}{2} ln x^{2} - 16 x - 64 - \frac{1}{2 2} (ln \frac{1}{8 2} x - \frac{1}{2} + 1 - ln \frac{1}{8 2} x - \frac{1}{2} - 1) + 2 (\frac{1}{2} ln \frac{1}{8 2} x - \frac{1}{2} + 1 - \frac{1}{2} ln \frac{1}{8 2} x - \frac{1}{2} - 1))) : - \frac{1}{8 2} ln \frac{1}{8 2} (x - 8) + 1 + \frac{1}{8 2} ln \frac{1}{8 2} (x - 8) - 1 + \frac{1}{64} ln ∣ x ∣ - \frac{1}{128 2} (2 ln x^{2} - 16 x - 64 - ln \frac{1}{8 2} x - \frac{1}{2} + 1 + ln \frac{1}{8 2} x - \frac{1}{2} - 1 + 2 ln \frac{x}{8 2} - \frac{1}{2} + 1 - 2 ln \frac{x}{8 2} - \frac{1}{2} - 1)

= - \frac{1}{8 2} ln \frac{1}{8 2} (x - 8) + 1 + \frac{1}{8 2} ln \frac{1}{8 2} (x - 8) - 1 + \frac{1}{64} ln ∣ x ∣ - \frac{1}{128 2} (2 ln x^{2} - 16 x - 64 - ln \frac{1}{8 2} x - \frac{1}{2} + 1 + ln \frac{1}{8 2} x - \frac{1}{2} - 1 + 2 ln \frac{x}{8 2} - \frac{1}{2} + 1 - 2 ln \frac{x}{8 2} - \frac{1}{2} - 1)

Add a constant to the solution

= - \frac{1}{8 2} ln \frac{1}{8 2} (x - 8) + 1 + \frac{1}{8 2} ln \frac{1}{8 2} (x - 8) - 1 + \frac{1}{64} ln ∣ x ∣ - \frac{1}{128 2} (2 ln x^{2} - 16 x - 64 - ln \frac{1}{8 2} x - \frac{1}{2} + 1 + ln \frac{1}{8 2} x - \frac{1}{2} - 1 + 2 ln \frac{x}{8 2} - \frac{1}{2} + 1 - 2 ln \frac{x}{8 2} - \frac{1}{2} - 1) + C

What is the integral of (2x-1)/(x^3-16x^2-64x) ?
The integral of (2x-1)/(x^3-16x^2-64x) is