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

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

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

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

x - \frac{3}{4} ln ∣ x + 1 ∣ + \frac{11}{4} ln ∣ x - 3 ∣ + C

Solution steps

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

Expand

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

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

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

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

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

Simplify

x - \frac{1}{4} ln ∣ x + 1 ∣ + \frac{9}{4} ln ∣ x - 3 ∣ + 2 (- \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 3 ∣) : x - \frac{3}{4} ln ∣ x + 1 ∣ + \frac{11}{4} ln ∣ x - 3 ∣

= x - \frac{3}{4} ln ∣ x + 1 ∣ + \frac{11}{4} ln ∣ x - 3 ∣

Add a constant to the solution

= x - \frac{3}{4} ln ∣ x + 1 ∣ + \frac{11}{4} ln ∣ x - 3 ∣ + C

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