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

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

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

\int \frac{4}{( x + 1 ) ( x + 3 )} d x

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

Solution steps

\int \frac{4}{( x + 1 ) ( 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 + 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 )}

= 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

x \to \infty lim (\frac{8 ^{x}}{9 ^{x + 1}})

\int \frac{32}{1 - cos ( 4 x )} d x

\int_{0}^{4} \frac{x}{x ^{2} - 4} d x

\frac{\partial}{\partial x} (r (θ, x) cos (θ))

x \to 3 - lim (\frac{3}{x ^{2} - 9})

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