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

The integral of (2x)/((x-1)(x+3)^2) is 2(1/16 ln|x-1|-1/16 ln|x+3|-3/(4(x+3)))+C

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

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

2 (\frac{1}{16} ln ∣ x - 1 ∣ - \frac{1}{16} ln ∣ x + 3 ∣ - \frac{3}{4 ( x + 3 )}) + C

Solution steps

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

Take the partial fraction of

\frac{x}{( x - 1 ) ( x + 3 ) ^{2}} : \frac{1}{16 ( x - 1 )} - \frac{1}{16 ( x + 3 )} + \frac{3}{4 ( x + 3 ) ^{2}}

= 2 \cdot \int \frac{1}{16 ( x - 1 )} - \frac{1}{16 ( x + 3 )} + \frac{3}{4 ( x + 3 ) ^{2}} 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}{16 ( x - 1 )} d x - \int \frac{1}{16 ( x + 3 )} d x + \int \frac{3}{4 ( x + 3 ) ^{2}} d x)

\int \frac{1}{16 ( x - 1 )} d x = \frac{1}{16} ln ∣ x - 1 ∣

\int \frac{1}{16 ( x + 3 )} d x = \frac{1}{16} ln ∣ x + 3 ∣

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

= 2 (\frac{1}{16} ln ∣ x - 1 ∣ - \frac{1}{16} ln ∣ x + 3 ∣ - \frac{3}{4 ( x + 3 )})

Add a constant to the solution

= 2 (\frac{1}{16} ln ∣ x - 1 ∣ - \frac{1}{16} ln ∣ x + 3 ∣ - \frac{3}{4 ( x + 3 )}) + C

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What is the integral of (2x)/((x-1)(x+3)^2) ?
The integral of (2x)/((x-1)(x+3)^2) is 2(1/16 ln|x-1|-1/16 ln|x+3|-3/(4(x+3)))+C