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

The integral of (x^2+x-5)/(2x^3+3x^2-2x) is 5/2 ln|x|-17/10 ln|2x-1|-3/10 ln|x+2|+C

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

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

\frac{5}{2} ln ∣ x ∣ - \frac{17}{10} ln ∣ 2 x - 1 ∣ - \frac{3}{10} ln ∣ x + 2 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{x ^{2} + x - 5}{2 x ^{3} + 3 x ^{2} - 2 x} : \frac{5}{2 x} - \frac{17}{5 ( 2 x - 1 )} - \frac{3}{10 ( x + 2 )}

= \int \frac{5}{2 x} - \frac{17}{5 ( 2 x - 1 )} - \frac{3}{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{17}{5 ( 2 x - 1 )} d x - \int \frac{3}{10 ( x + 2 )} d x

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

\int \frac{17}{5 ( 2 x - 1 )} d x = \frac{17}{10} ln ∣ 2 x - 1 ∣

\int \frac{3}{10 ( x + 2 )} d x = \frac{3}{10} ln ∣ x + 2 ∣

= \frac{5}{2} ln ∣ x ∣ - \frac{17}{10} ln ∣ 2 x - 1 ∣ - \frac{3}{10} ln ∣ x + 2 ∣

Add a constant to the solution

= \frac{5}{2} ln ∣ x ∣ - \frac{17}{10} ln ∣ 2 x - 1 ∣ - \frac{3}{10} ln ∣ x + 2 ∣ + C

\int cos (ln (2 x)) d x

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

\frac{d}{d x} (arccot (\frac{x}{4}))

\int \frac{y ^{3}}{( 2 + 3 y ^{4} ) ^{3}} d y

sin (tan (3 x))

What is the integral of (x^2+x-5)/(2x^3+3x^2-2x) ?
The integral of (x^2+x-5)/(2x^3+3x^2-2x) is 5/2 ln|x|-17/10 ln|2x-1|-3/10 ln|x+2|+C