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

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

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

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

2 (23 - \frac{4}{3})

Decimal

4.26153 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{3} 2 (u^{2} + 1) d u

= \int_{1}^{3} 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_{1}^{3} 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_{1}^{3} u^{2} d u + \int_{1}^{3} 1 d u)

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

\int_{1}^{3} 1 d u = 3 - 1

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

Simplify

2 (3 - \frac{1}{3} + 3 - 1) : 2 (23 - \frac{4}{3})

= 2 (23 - \frac{4}{3})

\int_{1}^{- 1} (3 r + 2)^{2} d r

\int_{0}^{4} (7 x^{2} - 8 x + 7) d x

\int_{- 1}^{1} 1 - \frac{8}{9 - x ^{2}} d x

\int_{1}^{7} \frac{2}{x} d x

\int_{0}^{- 1} 1

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