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

The integral of (x^2-3)/((x^2-4)(2x+1)) is 1/12 ln|x+2|+1/20 ln|x-2|+11/30 ln|2x+1|+C

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

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

\frac{1}{12} ln ∣ x + 2 ∣ + \frac{1}{20} ln ∣ x - 2 ∣ + \frac{11}{30} ln ∣ 2 x + 1 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{x ^{2} - 3}{( x ^{2} - 4 ) ( 2 x + 1 )} : \frac{1}{12 ( x + 2 )} + \frac{1}{20 ( x - 2 )} + \frac{11}{15 ( 2 x + 1 )}

= \int \frac{1}{12 ( x + 2 )} + \frac{1}{20 ( x - 2 )} + \frac{11}{15 ( 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{1}{12 ( x + 2 )} d x + \int \frac{1}{20 ( x - 2 )} d x + \int \frac{11}{15 ( 2 x + 1 )} d x

\int \frac{1}{12 ( x + 2 )} d x = \frac{1}{12} ln ∣ x + 2 ∣

\int \frac{1}{20 ( x - 2 )} d x = \frac{1}{20} ln ∣ x - 2 ∣

\int \frac{11}{15 ( 2 x + 1 )} d x = \frac{11}{30} ln ∣ 2 x + 1 ∣

= \frac{1}{12} ln ∣ x + 2 ∣ + \frac{1}{20} ln ∣ x - 2 ∣ + \frac{11}{30} ln ∣ 2 x + 1 ∣

Add a constant to the solution

= \frac{1}{12} ln ∣ x + 2 ∣ + \frac{1}{20} ln ∣ x - 2 ∣ + \frac{11}{30} ln ∣ 2 x + 1 ∣ + C

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What is the integral of (x^2-3)/((x^2-4)(2x+1)) ?
The integral of (x^2-3)/((x^2-4)(2x+1)) is 1/12 ln|x+2|+1/20 ln|x-2|+11/30 ln|2x+1|+C