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

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

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

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

ln ∣ x + 1 ∣ - ln ∣ x + 3 ∣ + C

Solution steps

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

Take the constant out:

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

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

Take the partial fraction of

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

= 2 \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

= 2 (\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 ∣

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

Simplify

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

= ln ∣ x + 1 ∣ - ln ∣ x + 3 ∣

Add a constant to the solution

= ln ∣ x + 1 ∣ - ln ∣ x + 3 ∣ + C

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

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

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

\int ln^{2} (x + 1 + x^{2}) d x

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