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

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

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

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

2

Solution steps

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

Apply u-substitution

: \int_{0}^{1} 2 d u

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

Integral of a constant:

\int a d x = a x

= [2 u]_{0}^{1}

Compute the boundaries:

2

= 2

\int \frac{16}{x ^{2} - 1} d x

x \frac{d y}{d x} - (1 + x) y = x y^{2}

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

\frac{d}{d x} (\frac{cos ( x )}{x ^{4}})

\frac{d}{d x} (\frac{y ^{2}}{x})

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