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

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

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

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

- \frac{1}{4} x + \frac{1}{32} ln ∣ 2 x + 1 ∣ - \frac{9}{32} ln ∣ 2 x - 3 ∣ + C

Solution steps

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

Take the partial fraction of

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

= \int - \frac{1}{4} + \frac{1}{16 ( 2 x + 1 )} - \frac{9}{16 ( 2 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}{4} d x + \int \frac{1}{16 ( 2 x + 1 )} d x - \int \frac{9}{16 ( 2 x - 3 )} d x

\int \frac{1}{4} d x = \frac{1}{4} x

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

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

= - \frac{1}{4} x + \frac{1}{32} ln ∣ 2 x + 1 ∣ - \frac{9}{32} ln ∣ 2 x - 3 ∣

Add a constant to the solution

= - \frac{1}{4} x + \frac{1}{32} ln ∣ 2 x + 1 ∣ - \frac{9}{32} ln ∣ 2 x - 3 ∣ + C

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