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

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

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

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

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

Solution steps

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

Expand

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

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

Simplify

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

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

Add a constant to the solution

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

\frac{d x}{d t} = k + x r

f (x) = x^{2} cos^{4} (x)

\int 4 t^{3} d t

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

\frac{d}{d x} (x^{4} + 8 x^{2} - x)

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