What is the integral of (2x-5)/(x^2-5x+2) ?

The integral of (2x-5)/(x^2-5x+2) is ln|4x^2-20x+8|+C

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integral of (2x-5)/(x^2-5x+2)

\int \frac{2 x - 5}{x ^{2} - 5 x + 2} d x

ln 4 x^{2} - 20 x + 8 + C

Solution steps

\int \frac{2 x - 5}{x ^{2} - 5 x + 2} d x

Expand

\frac{2 x - 5}{x ^{2} - 5 x + 2} : \frac{2 x}{x ^{2} - 5 x + 2} - \frac{5}{x ^{2} - 5 x + 2}

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 ^{2} - 5 x + 2} d x - \int \frac{5}{x ^{2} - 5 x + 2} d x

\int \frac{2 x}{x ^{2} - 5 x + 2} d x = ln 4 x^{2} - 20 x + 8 - \frac{5}{17} (ln \frac{2 x - 5}{17} + 1 - ln \frac{2 x - 5}{17} - 1)

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

= ln 4 x^{2} - 20 x + 8 - \frac{5}{17} (ln \frac{2 x - 5}{17} + 1 - ln \frac{2 x - 5}{17} - 1) - (- \frac{5}{17} (ln \frac{2 x - 5}{17} + 1 - ln \frac{2 x - 5}{17} - 1))

Simplify

ln 4 x^{2} - 20 x + 8 - \frac{5}{17} (ln \frac{2 x - 5}{17} + 1 - ln \frac{2 x - 5}{17} - 1) - (- \frac{5}{17} (ln \frac{2 x - 5}{17} + 1 - ln \frac{2 x - 5}{17} - 1)) : ln 4 x^{2} - 20 x + 8

= ln 4 x^{2} - 20 x + 8

Add a constant to the solution

= ln 4 x^{2} - 20 x + 8 + C

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\int \frac{x ^{2}}{x ^{2} - 81} d x

\int \frac{1}{x ( a x + b )} d x

What is the integral of (2x-5)/(x^2-5x+2) ?
The integral of (2x-5)/(x^2-5x+2) is ln|4x^2-20x+8|+C