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

The integral of (x^2+6x-5)/(2x^3+3x^2-2x) is 5/2 ln|x|-7/10 ln|2x-1|-13/10 ln|x+2|+C

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

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

\frac{5}{2} ln ∣ x ∣ - \frac{7}{10} ln ∣ 2 x - 1 ∣ - \frac{13}{10} ln ∣ x + 2 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{x ^{2} + 6 x - 5}{2 x ^{3} + 3 x ^{2} - 2 x} : \frac{5}{2 x} - \frac{7}{5 ( 2 x - 1 )} - \frac{13}{10 ( x + 2 )}

= \int \frac{5}{2 x} - \frac{7}{5 ( 2 x - 1 )} - \frac{13}{10 ( x + 2 )} d 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{5}{2 x} d x - \int \frac{7}{5 ( 2 x - 1 )} d x - \int \frac{13}{10 ( x + 2 )} d x

\int \frac{5}{2 x} d x = \frac{5}{2} ln ∣ x ∣

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

\int \frac{13}{10 ( x + 2 )} d x = \frac{13}{10} ln ∣ x + 2 ∣

= \frac{5}{2} ln ∣ x ∣ - \frac{7}{10} ln ∣ 2 x - 1 ∣ - \frac{13}{10} ln ∣ x + 2 ∣

Add a constant to the solution

= \frac{5}{2} ln ∣ x ∣ - \frac{7}{10} ln ∣ 2 x - 1 ∣ - \frac{13}{10} ln ∣ x + 2 ∣ + C

\frac{s + 1}{s ^{2} + 25}

\int 10 x e^{x^{2}} d x

f (x) = 1 + 2 e^{3 x}

8 (x - \frac{1}{x})^{2}

\frac{d ^{2} y}{d x ^{2}} y + 81 y = 0

What is the integral of (x^2+6x-5)/(2x^3+3x^2-2x) ?
The integral of (x^2+6x-5)/(2x^3+3x^2-2x) is 5/2 ln|x|-7/10 ln|2x-1|-13/10 ln|x+2|+C