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

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

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

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

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

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

\int \frac{1}{4 ( x - 1 )} d x = \frac{1}{4} ln ∣ x - 1 ∣

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

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

Simplify

4 (\frac{1}{4} ln ∣ x - 1 ∣ - \frac{1}{4} 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

f (x) = x^{2} x^{3} + 1, at x = 5

\frac{4 s}{( s ^{2} + 16 )}

x \to 0 lim (18 - x^{2})

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

x \to \infty lim (\frac{8}{x !})

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