Help with this integral please!

Question

anonymous · Answer

$$\int\limits_{a}^{b}\frac{ ax }{ \sqrt{x-1}}$$

solomonzelman · Answer

$\large\color{slate}{\displaystyle\int\limits_{a}^{b}\frac{ax}{\sqrt{x-1}}dx}$

solomonzelman · Answer

set   $u~~~=~~x-1$
        $du~=~~dx$

when,    $x=b$             then,    $u=b-1$
when,    $x=a$             then,    $u=a-1$

instead of   $ax$             you will have    $a(u+1)$
$($because when $u=x-1$ then $u+1=x$$~~)$

$\large\color{slate}{\displaystyle\int\limits_{b-1}^{a-1}\frac{a(u+1)}{\sqrt{u}}~du}$

solomonzelman · Answer

wrote the limits the opposite, sorry, I fixed it now.

$\large\color{slate}{\displaystyle\int\limits_{a-1}^{b-1}\frac{a(u+1)}{\sqrt{u}}~du}$

$\large\color{slate}{\displaystyle\int\limits_{a-1}^{b-1}\frac{au+a}{\sqrt{u}}~du}$

$\large\color{slate}{\displaystyle\int\limits_{a-1}^{b-1}\frac{au}{\sqrt{u}}+\frac{a}{\sqrt{u}}~du}$

$\large\color{slate}{\displaystyle\int\limits_{a-1}^{b-1}a(u)^{1/2} +a(u)^{-1/2} ~du}$

solomonzelman · Answer

$\large\color{slate}{\displaystyle \frac{2a}{3}(u)^{3/2} +2a(u)^{1/2} {\Huge |}^{b-1}_{a-1}}$

and so forth

anonymous · Answer

Thank you so much!

solomonzelman · Answer

yw