What is the integral of (x^2)/((x+1)(2x+3)) ?

The integral of (x^2)/((x+1)(2x+3)) is 1/2 x-5/8 ln(8|2x^2+5x+3|)+13/8 ln(4|x+1|)-13/8 ln(2|2x+3|)+C

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integral of (x^2)/((x+1)(2x+3))

integral

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

\frac{1}{2} x - \frac{5}{8} ln (8 2 x^{2} + 5 x + 3) + \frac{13}{8} ln (4 ∣ x + 1 ∣) - \frac{13}{8} ln (2 ∣ 2 x + 3 ∣) + C

Solution steps

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

Long division

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

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

\int \frac{1}{2} d x = \frac{1}{2} x

\int \frac{- 5 x - 3}{2 ( 2 x ^{2} + 5 x + 3 )} d x = - \frac{5}{8} ln (8 2 x^{2} + 5 x + 3) + \frac{13}{8} ln (4 ∣ x + 1 ∣) - \frac{13}{8} ln (2 ∣ 2 x + 3 ∣)

= \frac{1}{2} x - \frac{5}{8} ln (8 2 x^{2} + 5 x + 3) + \frac{13}{8} ln (4 ∣ x + 1 ∣) - \frac{13}{8} ln (2 ∣ 2 x + 3 ∣)

Add a constant to the solution

= \frac{1}{2} x - \frac{5}{8} ln (8 2 x^{2} + 5 x + 3) + \frac{13}{8} ln (4 ∣ x + 1 ∣) - \frac{13}{8} ln (2 ∣ 2 x + 3 ∣) + C

\frac{x}{( x - 2 ) ^{\frac{3}{2}}}

(e^{x})^{4}

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

\frac{x}{( 5 x + 2 ) ^{3}}

\frac{1}{( x ^{2} + 4 )}

What is the integral of (x^2)/((x+1)(2x+3)) ?
The integral of (x^2)/((x+1)(2x+3)) is 1/2 x-5/8 ln(8|2x^2+5x+3|)+13/8 ln(4|x+1|)-13/8 ln(2|2x+3|)+C