What is the integral of (11x^2+5x-4)/(2x^3-x^2) ?

The integral of (11x^2+5x-4)/(2x^3-x^2) is 5/2 ln|2x-1|+3ln|x|-4/x+C

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integral of (11x^2+5x-4)/(2x^3-x^2)

\int \frac{11 x ^{2} + 5 x - 4}{2 x ^{3} - x ^{2}} d x

\frac{5}{2} ln ∣ 2 x - 1 ∣ + 3 ln ∣ x ∣ - \frac{4}{x} + C

Solution steps

\int \frac{11 x ^{2} + 5 x - 4}{2 x ^{3} - x ^{2}} d x

Expand

\frac{11 x ^{2} + 5 x - 4}{2 x ^{3} - x ^{2}} : \frac{11 x ^{2}}{2 x ^{3} - x ^{2}} + \frac{5 x}{2 x ^{3} - x ^{2}} - \frac{4}{2 x ^{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}}{2 x ^{3} - x ^{2}} d x + \int \frac{5 x}{2 x ^{3} - x ^{2}} d x - \int \frac{4}{2 x ^{3} - x ^{2}} d x

\int \frac{11 x ^{2}}{2 x ^{3} - x ^{2}} d x = \frac{11}{2} ln ∣ 2 x - 1 ∣

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

\int \frac{4}{2 x ^{3} - x ^{2}} d x = 4 (- 2 ln ∣ x ∣ + \frac{1}{x} + 2 ln ∣ 2 x - 1 ∣)

= \frac{11}{2} ln ∣ 2 x - 1 ∣ + 5 (- ln ∣ x ∣ + ln ∣ 2 x - 1 ∣) - 4 (- 2 ln ∣ x ∣ + \frac{1}{x} + 2 ln ∣ 2 x - 1 ∣)

Simplify

\frac{11}{2} ln ∣ 2 x - 1 ∣ + 5 (- ln ∣ x ∣ + ln ∣ 2 x - 1 ∣) - 4 (- 2 ln ∣ x ∣ + \frac{1}{x} + 2 ln ∣ 2 x - 1 ∣) : \frac{5}{2} ln ∣ 2 x - 1 ∣ + 3 ln ∣ x ∣ - \frac{4}{x}

= \frac{5}{2} ln ∣ 2 x - 1 ∣ + 3 ln ∣ x ∣ - \frac{4}{x}

Add a constant to the solution

= \frac{5}{2} ln ∣ 2 x - 1 ∣ + 3 ln ∣ x ∣ - \frac{4}{x} + C

What is the integral of (11x^2+5x-4)/(2x^3-x^2) ?
The integral of (11x^2+5x-4)/(2x^3-x^2) is 5/2 ln|2x-1|+3ln|x|-4/x+C