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

The integral of (x^2)/((2x+3)(4x^2-1)) is 9/64 ln|2x+3|-1/32 ln|2x+1|+1/64 ln|2x-1|+C

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integral of (x^2)/((2x+3)(4x^2-1))

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

\frac{9}{64} ln ∣ 2 x + 3 ∣ - \frac{1}{32} ln ∣ 2 x + 1 ∣ + \frac{1}{64} ln ∣ 2 x - 1 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{x ^{2}}{( 2 x + 3 ) ( 4 x ^{2} - 1 )} : \frac{9}{32 ( 2 x + 3 )} - \frac{1}{16 ( 2 x + 1 )} + \frac{1}{32 ( 2 x - 1 )}

= \int \frac{9}{32 ( 2 x + 3 )} - \frac{1}{16 ( 2 x + 1 )} + \frac{1}{32 ( 2 x - 1 )} 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{9}{32 ( 2 x + 3 )} d x - \int \frac{1}{16 ( 2 x + 1 )} d x + \int \frac{1}{32 ( 2 x - 1 )} d x

\int \frac{9}{32 ( 2 x + 3 )} d x = \frac{9}{64} ln ∣ 2 x + 3 ∣

\int \frac{1}{16 ( 2 x + 1 )} d x = \frac{1}{32} ln ∣ 2 x + 1 ∣

\int \frac{1}{32 ( 2 x - 1 )} d x = \frac{1}{64} ln ∣ 2 x - 1 ∣

= \frac{9}{64} ln ∣ 2 x + 3 ∣ - \frac{1}{32} ln ∣ 2 x + 1 ∣ + \frac{1}{64} ln ∣ 2 x - 1 ∣

Add a constant to the solution

= \frac{9}{64} ln ∣ 2 x + 3 ∣ - \frac{1}{32} ln ∣ 2 x + 1 ∣ + \frac{1}{64} ln ∣ 2 x - 1 ∣ + C

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What is the integral of (x^2)/((2x+3)(4x^2-1)) ?
The integral of (x^2)/((2x+3)(4x^2-1)) is 9/64 ln|2x+3|-1/32 ln|2x+1|+1/64 ln|2x-1|+C