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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{2 u - 3}{u ( u - 4 ) ^{2}} d u

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

Expand

\frac{2 u - 3}{u ( u - 4 ) ^{2}} : \frac{2 u}{u ( u - 4 ) ^{2}} - \frac{3}{u ( u - 4 ) ^{2}}

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{2 u}{u ( u - 4 ) ^{2}} d u - \int \frac{3}{u ( u - 4 ) ^{2}} d u

\int \frac{2 u}{u ( u - 4 ) ^{2}} d u = - \frac{2}{u - 4}

\int \frac{3}{u ( u - 4 ) ^{2}} d u = 3 (\frac{1}{16} ln ∣ u ∣ - \frac{1}{16} ln ∣ u - 4 ∣ - \frac{1}{4 ( u - 4 )})

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

Substitute back

u = x + 3

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

Simplify

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

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

Add a constant to the solution

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

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