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

The integral of (x^2+1)/((x^2-4)(x+2)) is 11/16 ln|x+2|+5/(4(x+2))+5/16 ln|x-2|+C

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

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

\frac{11}{16} ln ∣ x + 2 ∣ + \frac{5}{4 ( x + 2 )} + \frac{5}{16} ln ∣ x - 2 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{x ^{2} + 1}{( x ^{2} - 4 ) ( x + 2 )} : \frac{11}{16 ( x + 2 )} - \frac{5}{4 ( x + 2 ) ^{2}} + \frac{5}{16 ( x - 2 )}

= \int \frac{11}{16 ( x + 2 )} - \frac{5}{4 ( x + 2 ) ^{2}} + \frac{5}{16 ( x - 2 )} 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{11}{16 ( x + 2 )} d x - \int \frac{5}{4 ( x + 2 ) ^{2}} d x + \int \frac{5}{16 ( x - 2 )} d x

\int \frac{11}{16 ( x + 2 )} d x = \frac{11}{16} ln ∣ x + 2 ∣

\int \frac{5}{4 ( x + 2 ) ^{2}} d x = - \frac{5}{4 ( x + 2 )}

\int \frac{5}{16 ( x - 2 )} d x = \frac{5}{16} ln ∣ x - 2 ∣

= \frac{11}{16} ln ∣ x + 2 ∣ - (- \frac{5}{4 ( x + 2 )}) + \frac{5}{16} ln ∣ x - 2 ∣

Simplify

= \frac{11}{16} ln ∣ x + 2 ∣ + \frac{5}{4 ( x + 2 )} + \frac{5}{16} ln ∣ x - 2 ∣

Add a constant to the solution

= \frac{11}{16} ln ∣ x + 2 ∣ + \frac{5}{4 ( x + 2 )} + \frac{5}{16} ln ∣ x - 2 ∣ + C

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What is the integral of (x^2+1)/((x^2-4)(x+2)) ?
The integral of (x^2+1)/((x^2-4)(x+2)) is 11/16 ln|x+2|+5/(4(x+2))+5/16 ln|x-2|+C