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

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

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

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \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 )}

= 4 \cdot \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

= 4 (\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 ∣

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

Add a constant to the solution

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

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