# How do you find the integral of int dx/(sqrt(2x-1) from 1/2 to 2?

Oct 14, 2015

Try the substitution $u = 2 x - 1$ in hopes of getting $\int {u}^{- \frac{1}{2}} \mathrm{du}$.

#### Explanation:

With $u = 2 x - 1$, we get $\mathrm{du} = 2 \mathrm{dx}$ so the indefinite integral becomes

$\int {\left(2 x - 1\right)}^{- \frac{1}{2}} \mathrm{dx} = \frac{1}{2} \int {u}^{- \frac{1}{2}} \mathrm{du}$

The limits of integration change from $x = \frac{1}{2}$ to $u = 2 \left(\frac{1}{2}\right) - 1 = 0$
and from $x = 2$ to $u = 2 \left(2\right) - 1 = 3$

The new problem is to evaluate

$\frac{1}{2} {\int}_{0}^{3} {u}^{- \frac{1}{2}} \mathrm{du}$

$= {\left[{u}^{\frac{1}{2}}\right]}_{0}^{3} = \sqrt{3}$