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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

= 2 \cdot \int \frac{1}{4 ( x - 1 )} + \frac{3}{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

= 2 (\int \frac{1}{4 ( x - 1 )} d x + \int \frac{3}{4 ( x + 3 )} d x)

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

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

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

Add a constant to the solution

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

x^{6} e^{x}

f (x) = 4 x^{2} + 2 x, (- 3, 30)

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

(x - 3) 4 - x

\frac{d}{d x} ((1 - x^{3})^{\frac{1}{2}})

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