Question

Evaluate ${\int }_1^5\frac{dx}{\sqrt{2x-1}}$.

Answer 1

Let $I={\int }_1^5\frac{dx}{\sqrt{2x-1}}$.

Substitute $2x-1=t^2\Rightarrow dx=tdt$

Also upper limit $\sqrt{2\left(5\right)-1}=3$

and lower limit $=\sqrt{2\left(1\right)-1}=1$

Thus

$I={\int }_1^3\frac{tdt}{t}={\int }_1^{3}dt=[t{]}_1^3=3-1=2$.