What is the integral from 3 to 4 of (x^2-1)/(x-1) ?

The integral from 3 to 4 of (x^2-1)/(x-1) is 9/2

integral from 3 to 4 of (x^2-1)/(x-1)

\int_{3}^{4} \frac{x ^{2} - 1}{x - 1} d x

\frac{9}{2}

Decimal

4.5

Solution steps

\int_{3}^{4} \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_{3}^{4} \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_{3}^{4} 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_{3}^{4} x d x + \int_{3}^{4} 1 d x

\int_{3}^{4} x d x = \frac{7}{2}

\int_{3}^{4} 1 d x = 1

= \frac{7}{2} + 1

Simplify

\frac{7}{2} + 1 : \frac{9}{2}

= \frac{9}{2}

\int_{- 1}^{1} \frac{x ^{2}}{x ^{3} - 1} d x

\int_{0}^{π} sin^{2} (n) x d x

\int_{x}^{y} cos (e^{t}) d t

\int_{0}^{2 π} 3 sin (\frac{x}{2}) d x

\int_{1}^{4} \frac{x ^{2}}{x ^{2} + 4} d x

What is the integral from 3 to 4 of (x^2-1)/(x-1) ?
The integral from 3 to 4 of (x^2-1)/(x-1) is 9/2