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

The integral of (x^2-2x-3)/((x-1)(x^2+2x+2)) is -4/5 ln|x-1|+1/10 (9ln|x^2+2x+2|-4arctan(x+1))+C

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

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

- \frac{4}{5} ln ∣ x - 1 ∣ + \frac{1}{10} (9 ln x^{2} + 2 x + 2 - 4 arctan (x + 1)) + C

Solution steps

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

Take the partial fraction of

\frac{x ^{2} - 2 x - 3}{( x - 1 ) ( x ^{2} + 2 x + 2 )} : - \frac{4}{5 ( x - 1 )} + \frac{9 x + 7}{5 ( x ^{2} + 2 x + 2 )}

= \int - \frac{4}{5 ( x - 1 )} + \frac{9 x + 7}{5 ( x ^{2} + 2 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{4}{5 ( x - 1 )} d x + \int \frac{9 x + 7}{5 ( x ^{2} + 2 x + 2 )} d x

\int \frac{4}{5 ( x - 1 )} d x = \frac{4}{5} ln ∣ x - 1 ∣

\int \frac{9 x + 7}{5 ( x ^{2} + 2 x + 2 )} d x = \frac{1}{10} (9 ln x^{2} + 2 x + 2 - 4 arctan (x + 1))

= - \frac{4}{5} ln ∣ x - 1 ∣ + \frac{1}{10} (9 ln x^{2} + 2 x + 2 - 4 arctan (x + 1))

Add a constant to the solution

= - \frac{4}{5} ln ∣ x - 1 ∣ + \frac{1}{10} (9 ln x^{2} + 2 x + 2 - 4 arctan (x + 1)) + C

What is the integral of (x^2-2x-3)/((x-1)(x^2+2x+2)) ?
The integral of (x^2-2x-3)/((x-1)(x^2+2x+2)) is -4/5 ln|x-1|+1/10 (9ln|x^2+2x+2|-4arctan(x+1))+C