Answered By Majid B

 $\int\frac{x}{\left(x-1\right)^{\frac{1}{2}}}dx$∫x(x−1)¹²  dx $=$= $\int u.dv=u.v-\int v.du$∫u.‍dv=u.v−∫v.du

 $x=u$‍x=u    =>    $dx=du$dx=du 

$\frac{dx}{\left(x-1\right)^{\frac{1}{2}}}=dv$dx(x−1)¹²  =dv     =>    $2\left(x-1\right)^{\frac{1}{2}}=v$2(x−1)¹² =v 

        $\int\frac{x}{\left(x-1\right)^{\frac{1}{2}}}dx=x.\left(2\left(x-1\right)^{\frac{1}{2}}\right)-\int2\left(x-1\right)^{\frac{1}{2}}.dx=2x\left(x-1\right)^{\frac{1}{2}}-2\left(\frac{\left(x-1\right)^{\frac{3}{2}}}{\frac{3}{2}}\right)+C=2x\left(x-1\right)^{\frac{1}{2}}-\frac{4}{3}\left(x-1\right)^{\frac{3}{2}}+C=\frac{1}{3}\left(x-1\right)^{\frac{1}{2}}\left(6x-4x+4\right)+C=\frac{1}{3}\left(x-1\right)^{\frac{1}{2}}\left(2x+4\right)+C=\frac{2}{3}\left(x-1\right)^{\frac{1}{2}}\left(x+2\right)+C$∫x(x−1)¹²  dx=x.(2(x−1)¹² )−∫2(x−1)¹² .dx=2x(x−1)¹² −2((x−1)³² 32  )+C=2x(x−1)¹² −43 (x−1)³² +C=13 (x−1)¹² (6x−4x+4)+C=13 (x−1)¹² (2x+4)+C=23 (x−1)¹² (x+2)+C‍    

Answered By Michael Z

1. Let u=x and dv= $\frac{1}{\sqrt{x-1}}$1√x−1  , then du=dx and v= $2\left(x-1\right)^{\frac{1}{2}}$2(x−1)¹²  [hint use substitution]

2.integration by parts formula is  $\int udv=uv-\int vdu$∫udv=uv−∫vdu 

3. substitute:   $\int\frac{x}{\sqrt{x-1}}=\left(x\right)\left(2\left(x-1\right)^{\frac{1}{2}}\right)-\int2\left(x-1\right)^{\frac{1}{2}}dx$∫x√x−1 =(x)(2(x−1‍)¹² )−∫2(x−1)¹² dx  

4. integrate:  $\int2\left(x-1\right)^{\frac{1}{2}}dx=\frac{4\left(x-1\right)^{\frac{3}{2}}}{3}+c$∫2(x−1)¹² dx=4(x−1)³² 3 +‍c  [hint use substitution]

5. solution is:   $\int\frac{x}{\sqrt{x-1}}=\left(x\right)\left(2\left(x-1\right)^{\frac{1}{2}}\right)-\frac{4\left(x-1\right)^{\frac{3}{2}}}{3}+c$∫x√x−1 =(x)(2(x−1‍)¹² )−4(x−1)³² 3 +c