What is the integral of (6x-5)/(3x^2-5x) ?

The integral of (6x-5)/(3x^2-5x) is 2ln|3x-5|+ln(6/5 |x|)-ln| 6/5 x-2|+C

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integral of (6x-5)/(3x^2-5x)

\int \frac{6 x - 5}{3 x ^{2} - 5 x} d x

2 ln ∣ 3 x - 5 ∣ + ln (\frac{6}{5} ∣ x ∣) - ln \frac{6}{5} x - 2 + C

Solution steps

\int \frac{6 x - 5}{3 x ^{2} - 5 x} d x

Expand

\frac{6 x - 5}{3 x ^{2} - 5 x} : \frac{6 x}{3 x ^{2} - 5 x} - \frac{5}{3 x ^{2} - 5 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{6 x}{3 x ^{2} - 5 x} d x - \int \frac{5}{3 x ^{2} - 5 x} d x

\int \frac{6 x}{3 x ^{2} - 5 x} d x = 2 ln ∣ 3 x - 5 ∣

\int \frac{5}{3 x ^{2} - 5 x} d x = - ln (\frac{6}{5} ∣ x ∣) + ln \frac{6}{5} x - 2

= 2 ln ∣ 3 x - 5 ∣ - (- ln (\frac{6}{5} ∣ x ∣) + ln \frac{6}{5} x - 2)

Simplify

= 2 ln ∣ 3 x - 5 ∣ + ln (\frac{6}{5} ∣ x ∣) - ln \frac{6}{5} x - 2

Add a constant to the solution

= 2 ln ∣ 3 x - 5 ∣ + ln (\frac{6}{5} ∣ x ∣) - ln \frac{6}{5} x - 2 + C

\int_{36}^{x} 1 + 9 t d t

\frac{d}{d x} (ln (x^{4} + 1))

\frac{d}{d x} (9 x^{2} sin (x) tan (x))

y^{^{''}} + 10 y^{^{'}} + 25 y = 5 e^{- 5 t}

n = 0 \sum \infty - x^{n}

What is the integral of (6x-5)/(3x^2-5x) ?
The integral of (6x-5)/(3x^2-5x) is 2ln|3x-5|+ln(6/5 |x|)-ln| 6/5 x-2|+C