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

The integral from 1 to 2 of x/(sqrt(x-1)) is 8/3

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integral from 1 to 2 of x/(sqrt(x-1))

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

\frac{8}{3}

Decimal

2.66666 \dots

Solution steps

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

Apply u-substitution

: \int_{0}^{1} 2 (u^{2} + 1) d u

= \int_{0}^{1} 2 (u^{2} + 1) d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int_{0}^{1} u^{2} + 1 d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 (\int_{0}^{1} u^{2} d u + \int_{0}^{1} 1 d u)

\int_{0}^{1} u^{2} d u = \frac{1}{3}

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

= 2 (\frac{1}{3} + 1)

Simplify

2 (\frac{1}{3} + 1) : \frac{8}{3}

= \frac{8}{3}

\int_{0}^{3} 2 sin (x) - e^{x} d x

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

\int_{0}^{x} cos (t^{2}) d t

\int_{0}^{\frac{π}{3}} cos (x) d x

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

What is the integral from 1 to 2 of x/(sqrt(x-1)) ?
The integral from 1 to 2 of x/(sqrt(x-1)) is 8/3