What is the integral from 0 to 1 of (x^2-1)/(x-1) ?

The integral from 0 to 1 of (x^2-1)/(x-1) is 3/2

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integral from 0 to 1 of (x^2-1)/(x-1)

\int_{0}^{1} \frac{x ^{2} - 1}{x - 1} d x

\frac{3}{2}

Decimal

1.5

Solution steps

\int_{0}^{1} \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_{0}^{1} \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_{0}^{1} 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_{0}^{1} x d x + \int_{0}^{1} 1 d x

\int_{0}^{1} x d x = \frac{1}{2}

\int_{0}^{1} 1 d x = 1

= \frac{1}{2} + 1

Simplify

\frac{1}{2} + 1 : \frac{3}{2}

= \frac{3}{2}

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What is the integral from 0 to 1 of (x^2-1)/(x-1) ?
The integral from 0 to 1 of (x^2-1)/(x-1) is 3/2