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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

Add a constant to the solution

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

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