What is the integral of (4x+5)/((x-1)(x+2)^2) ?

The integral of (4x+5)/((x-1)(x+2)^2) is ln|x-1|-ln|x+2|-1/(x+2)+C

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integral of (4x+5)/((x-1)(x+2)^2)

\int \frac{4 x + 5}{( x - 1 ) ( x + 2 ) ^{2}} d x

ln ∣ x - 1 ∣ - ln ∣ x + 2 ∣ - \frac{1}{x + 2} + C

Solution steps

\int \frac{4 x + 5}{( x - 1 ) ( x + 2 ) ^{2}} d x

Take the partial fraction of

\frac{4 x + 5}{( x - 1 ) ( x + 2 ) ^{2}} : \frac{1}{x - 1} - \frac{1}{x + 2} + \frac{1}{( x + 2 ) ^{2}}

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

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

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

\int \frac{1}{x + 2} d x = ln ∣ x + 2 ∣

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

= ln ∣ x - 1 ∣ - ln ∣ x + 2 ∣ - \frac{1}{x + 2}

Add a constant to the solution

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

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What is the integral of (4x+5)/((x-1)(x+2)^2) ?
The integral of (4x+5)/((x-1)(x+2)^2) is ln|x-1|-ln|x+2|-1/(x+2)+C