What is the integral of+(8x-3)/(12x^2-7x+1) ?

The integral of+(8x-3)/(12x^2-7x+1) is 8/3 ln|3x-1|+3ln|24x-6|-3ln|24x-8|-2ln|4x-1|-8/3+2+C

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integral of+(8x-3)/(12x^2-7x+1)

\int + \frac{8 x - 3}{12 x ^{2} - 7 x + 1} d x

\frac{8}{3} ln ∣ 3 x - 1 ∣ + 3 ln ∣ 24 x - 6 ∣ - 3 ln ∣ 24 x - 8 ∣ - 2 ln ∣ 4 x - 1 ∣ - \frac{8}{3} + 2 + C

Solution steps

\int \frac{8 x - 3}{12 x ^{2} - 7 x + 1} d x

Apply Integration By Parts

: (8 x - 3) (- ln ∣ 24 x - 6 ∣ + ln ∣ 24 x - 8 ∣) - \int 8 (- ln ∣ 24 x - 6 ∣ + ln ∣ 24 x - 8 ∣) d x

= (8 x - 3) (- ln ∣ 24 x - 6 ∣ + ln ∣ 24 x - 8 ∣) - \int 8 (- ln ∣ 24 x - 6 ∣ + ln ∣ 24 x - 8 ∣) d x

\int 8 (- ln ∣ 24 x - 6 ∣ + ln ∣ 24 x - 8 ∣) d x = 8 (- x ln ∣ 24 x - 6 ∣ + x ln ∣ 24 x - 8 ∣ + \frac{1}{4} ln ∣ 4 x - 1 ∣ - \frac{1}{3} ln ∣ 3 x - 1 ∣ - \frac{1}{4} + \frac{1}{3})

= (8 x - 3) (- ln ∣ 24 x - 6 ∣ + ln ∣ 24 x - 8 ∣) - 8 (- x ln ∣ 24 x - 6 ∣ + x ln ∣ 24 x - 8 ∣ + \frac{1}{4} ln ∣ 4 x - 1 ∣ - \frac{1}{3} ln ∣ 3 x - 1 ∣ - \frac{1}{4} + \frac{1}{3})

Simplify

(8 x - 3) (- ln ∣ 24 x - 6 ∣ + ln ∣ 24 x - 8 ∣) - 8 (- x ln ∣ 24 x - 6 ∣ + x ln ∣ 24 x - 8 ∣ + \frac{1}{4} ln ∣ 4 x - 1 ∣ - \frac{1}{3} ln ∣ 3 x - 1 ∣ - \frac{1}{4} + \frac{1}{3}) : \frac{8}{3} ln ∣ 3 x - 1 ∣ + 3 ln ∣ 24 x - 6 ∣ - 3 ln ∣ 24 x - 8 ∣ - 2 ln ∣ 4 x - 1 ∣ - \frac{8}{3} + 2

= \frac{8}{3} ln ∣ 3 x - 1 ∣ + 3 ln ∣ 24 x - 6 ∣ - 3 ln ∣ 24 x - 8 ∣ - 2 ln ∣ 4 x - 1 ∣ - \frac{8}{3} + 2

Add a constant to the solution

= \frac{8}{3} ln ∣ 3 x - 1 ∣ + 3 ln ∣ 24 x - 6 ∣ - 3 ln ∣ 24 x - 8 ∣ - 2 ln ∣ 4 x - 1 ∣ - \frac{8}{3} + 2 + C

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What is the integral of+(8x-3)/(12x^2-7x+1) ?
The integral of+(8x-3)/(12x^2-7x+1) is 8/3 ln|3x-1|+3ln|24x-6|-3ln|24x-8|-2ln|4x-1|-8/3+2+C