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

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

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

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

43

Solution steps

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

Apply u-substitution

: \int_{0}^{3} 2 (u^{2} + 1) d u

= \int_{0}^{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_{0}^{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_{0}^{3} u^{2} d u + \int_{0}^{3} 1 d u)

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

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

= 2 (3 + 3)

Simplify

= 43

\int_{0}^{2 - x} (2 - x - y) d y

\int_{0}^{1} (1 - 8 x^{3} + 16 x^{7}) d x

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