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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

Add a constant to the solution

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

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