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

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

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

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

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

Solution steps

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

Simplify

\frac{x ^{2} + x}{2 x ^{3} + 3 x ^{2}} : \frac{x + 1}{x ( 2 x + 3 )}

= \int \frac{x + 1}{x ( 2 x + 3 )} d x

Expand

\frac{x + 1}{x ( 2 x + 3 )} : \frac{x}{x ( 2 x + 3 )} + \frac{1}{x ( 2 x + 3 )}

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{x}{x ( 2 x + 3 )} d x + \int \frac{1}{x ( 2 x + 3 )} d x

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

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

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

Add a constant to the solution

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

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