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

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

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

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

- ln ∣ x + 1 ∣ + 2 ln ∣ x + 3 ∣ + C

Solution steps

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

Take the partial fraction of

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

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

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

\int \frac{2}{x + 3} d x = 2 ln ∣ x + 3 ∣

= - ln ∣ x + 1 ∣ + 2 ln ∣ x + 3 ∣

Add a constant to the solution

= - ln ∣ x + 1 ∣ + 2 ln ∣ x + 3 ∣ + C

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