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

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

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

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

\frac{x ^{2}}{2} + x + C

Solution steps

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

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

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

Simplify

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

= \int x + 1 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 x d x + \int 1 d x

\int x d x = \frac{x ^{2}}{2}

\int 1 d x = x

= \frac{x ^{2}}{2} + x

Add a constant to the solution

= \frac{x ^{2}}{2} + x + C

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