What is the integral of (31x-33)/(2x^2-6x) ?

The integral of (31x-33)/(2x^2-6x) is (31x-33)/(2(x-3))+1/2 (11ln|x|+20ln|x-3|-(60)/(x-3))+C

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integral of (31x-33)/(2x^2-6x)

\int \frac{31 x - 33}{2 x ^{2} - 6 x} d x

\frac{31 x - 33}{2 ( x - 3 )} + \frac{1}{2} (11 ln ∣ x ∣ + 20 ln ∣ x - 3 ∣ - \frac{60}{x - 3}) + C

Solution steps

\int \frac{31 x - 33}{2 x ^{2} - 6 x} d x

Apply Integration By Parts

: \frac{31 x - 33}{2 ( x - 3 )} - \int - \frac{31 x ^{2} - 66 x + 99}{2 x ( x - 3 ) ^{2}} d x

= \frac{31 x - 33}{2 ( x - 3 )} - \int - \frac{31 x ^{2} - 66 x + 99}{2 x ( x - 3 ) ^{2}} d x

\int - \frac{31 x ^{2} - 66 x + 99}{2 x ( x - 3 ) ^{2}} d x = - \frac{1}{2} (11 ln ∣ x ∣ + 20 ln ∣ x - 3 ∣ - \frac{60}{x - 3})

= \frac{31 x - 33}{2 ( x - 3 )} - (- \frac{1}{2} (11 ln ∣ x ∣ + 20 ln ∣ x - 3 ∣ - \frac{60}{x - 3}))

Simplify

= \frac{31 x - 33}{2 ( x - 3 )} + \frac{1}{2} (11 ln ∣ x ∣ + 20 ln ∣ x - 3 ∣ - \frac{60}{x - 3})

Add a constant to the solution

= \frac{31 x - 33}{2 ( x - 3 )} + \frac{1}{2} (11 ln ∣ x ∣ + 20 ln ∣ x - 3 ∣ - \frac{60}{x - 3}) + C

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What is the integral of (31x-33)/(2x^2-6x) ?
The integral of (31x-33)/(2x^2-6x) is (31x-33)/(2(x-3))+1/2 (11ln|x|+20ln|x-3|-(60)/(x-3))+C