How do you evaluate the integral $\int \frac{3 x^{2} - 4 x + 2}{x - 5}$ $int (3x^2-4x+2)/(x-5)$ ?

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Answer 1 · 2017-03-05T19:45:27

$= \frac{3 x^{2}}{2} + 11 x + 57 ln | x - 5 | + C$ $= (3x^2)/2 + 11x + 57ln |x-5| + C$

If you do the long division, you will see that:

$\frac{3 x^{2} - 4 x + 2}{x - 5} = 3 x + 11 + \frac{57}{x - 5}$ $(3x^2-4x+2)/(x-5) = 3x + 11 + 57/(x-5)$

And:

$int dx qquad ( 3x + 11 + 57/(x-5) )$

$= (3x^2)/2 + 11x + 57ln |x-5| + C$

How do you evaluate the integral int (3x^2-4x+2)/(x-5)∫3x2−4x+2x−5int (3x^2-4x+2)/(x-5)?