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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

\int \frac{5}{x - 1} d x = 5 ln ∣ x - 1 ∣

= \frac{( x - 1 ) ^{2}}{2} + 2 (x - 1) + ln ∣ x - 1 ∣ + 2 (x - 1 + ln ∣ x - 1 ∣) - 5 ln ∣ x - 1 ∣

Simplify

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

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

Add a constant to the solution

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

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