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

The integral of (x^2+1)/(x-2) is 1/2 (x-2)^2+4x-8+5ln|x-2|+C

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

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

\frac{1}{2} (x - 2)^{2} + 4 x - 8 + 5 ln ∣ x - 2 ∣ + C

Solution steps

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

Expand

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

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

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

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

Simplify

\frac{( x - 2 ) ^{2}}{2} + 4 (x - 2) + 4 ln ∣ x - 2 ∣ + ln ∣ x - 2 ∣ : \frac{1}{2} (x - 2)^{2} + 4 x - 8 + 5 ln ∣ x - 2 ∣

= \frac{1}{2} (x - 2)^{2} + 4 x - 8 + 5 ln ∣ x - 2 ∣

Add a constant to the solution

= \frac{1}{2} (x - 2)^{2} + 4 x - 8 + 5 ln ∣ x - 2 ∣ + C

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What is the integral of (x^2+1)/(x-2) ?
The integral of (x^2+1)/(x-2) is 1/2 (x-2)^2+4x-8+5ln|x-2|+C