What is the integral from 2 to 5 of 1/(sqrt(x-1)) ?

The integral from 2 to 5 of 1/(sqrt(x-1)) is 2

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integral from 2 to 5 of 1/(sqrt(x-1))

\int_{2}^{5} \frac{1}{x - 1} d x

2

Solution steps

\int_{2}^{5} \frac{1}{x - 1} d x

Apply u-substitution

: \int_{1}^{2} 2 d u

= \int_{1}^{2} 2 d u

Integral of a constant:

\int a d x = a x

= [2 u]_{1}^{2}

Compute the boundaries:

2

= 2

\int x^{2} (2 x^{3} + 4)^{5} d x

\int \frac{x}{k} d x

x^{2} + x^{3} 1

(4 t^{2} - t) (t^{3} - 8 t^{2} + 12)

y = sin (x), y = 2 sin^{2} (x), x = 0, x = \frac{π}{2}

What is the integral from 2 to 5 of 1/(sqrt(x-1)) ?
The integral from 2 to 5 of 1/(sqrt(x-1)) is 2