What is the integral of ((11x-7))/(2x^2-3x-2) ?

The integral of ((11x-7))/(2x^2-3x-2) is 5/2 ln|2x+1|+3ln|x-2|+C

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integral of ((11x-7))/(2x^2-3x-2)

\int \frac{( 11 x - 7 )}{2 x ^{2} - 3 x - 2} d x

\frac{5}{2} ln ∣ 2 x + 1 ∣ + 3 ln ∣ x - 2 ∣ + C

Solution steps

\int \frac{11 x - 7}{2 x ^{2} - 3 x - 2} d x

Expand

\frac{11 x - 7}{2 x ^{2} - 3 x - 2} : \frac{11 x}{2 x ^{2} - 3 x - 2} - \frac{7}{2 x ^{2} - 3 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{11 x}{2 x ^{2} - 3 x - 2} d x - \int \frac{7}{2 x ^{2} - 3 x - 2} d x

\int \frac{11 x}{2 x ^{2} - 3 x - 2} d x = 11 (\frac{1}{10} ln ∣ 2 x + 1 ∣ + \frac{2}{5} ln ∣ x - 2 ∣)

\int \frac{7}{2 x ^{2} - 3 x - 2} d x = 7 (- \frac{1}{5} ln ∣ 2 x + 1 ∣ + \frac{1}{5} ln ∣ x - 2 ∣)

= 11 (\frac{1}{10} ln ∣ 2 x + 1 ∣ + \frac{2}{5} ln ∣ x - 2 ∣) - 7 (- \frac{1}{5} ln ∣ 2 x + 1 ∣ + \frac{1}{5} ln ∣ x - 2 ∣)

Simplify

11 (\frac{1}{10} ln ∣ 2 x + 1 ∣ + \frac{2}{5} ln ∣ x - 2 ∣) - 7 (- \frac{1}{5} ln ∣ 2 x + 1 ∣ + \frac{1}{5} ln ∣ x - 2 ∣) : \frac{5}{2} ln ∣ 2 x + 1 ∣ + 3 ln ∣ x - 2 ∣

= \frac{5}{2} ln ∣ 2 x + 1 ∣ + 3 ln ∣ x - 2 ∣

Add a constant to the solution

= \frac{5}{2} ln ∣ 2 x + 1 ∣ + 3 ln ∣ x - 2 ∣ + C

\int \frac{e ^{x}}{e ^{2 x}} d x

\int \frac{16 x}{8 x ^{2} + 2} d x

\int \frac{2 sin ^{2} ( x )}{6} d x

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

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

What is the integral of ((11x-7))/(2x^2-3x-2) ?
The integral of ((11x-7))/(2x^2-3x-2) is 5/2 ln|2x+1|+3ln|x-2|+C