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

The integral of (2p+3)/((p-1)^2) is 2ln|p-1|-5/(p-1)+C

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

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

2 ln ∣ p - 1 ∣ - \frac{5}{p - 1} + C

Solution steps

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

Expand

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

\int \frac{2 p}{( p - 1 ) ^{2}} d p = 2 (ln ∣ p - 1 ∣ - \frac{1}{p - 1})

\int \frac{3}{( p - 1 ) ^{2}} d p = - \frac{3}{p - 1}

= 2 (ln ∣ p - 1 ∣ - \frac{1}{p - 1}) - \frac{3}{p - 1}

Simplify

2 (ln ∣ p - 1 ∣ - \frac{1}{p - 1}) - \frac{3}{p - 1} : 2 ln ∣ p - 1 ∣ - \frac{5}{p - 1}

= 2 ln ∣ p - 1 ∣ - \frac{5}{p - 1}

Add a constant to the solution

= 2 ln ∣ p - 1 ∣ - \frac{5}{p - 1} + C

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What is the integral of (2p+3)/((p-1)^2) ?
The integral of (2p+3)/((p-1)^2) is 2ln|p-1|-5/(p-1)+C