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

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

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

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

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

Solution steps

\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 )}

= \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

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

f (x) = x^{4} ln (x)

y = 7 x - x^{2} (1.6)

e^{x} - 2 x

\frac{d}{d x} (\frac{2 x ^{2} + 4 x + 6}{x})

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