- #1
ISU20CpreE
- 69
- 0
ok i got this problem:
[tex]\int\frac{\sqrt{x-2}} {x+1} \,dx[/tex]
then i have to let u=(x-2)^0.5 then i solve for x in terms of u
so i get x= u^2+2
dx=2u du
[tex]\int\frac{\sqrt{x-2}} {x+1} \,dx=\int\frac{u}{u^2+3}\,2u\,du[/tex]
where when i substitute i get:
[tex]\int\frac{2u^2} {u^2+3}\,du[/tex]. after that I am completely lost.
[tex]\int\frac{\sqrt{x-2}} {x+1} \,dx[/tex]
then i have to let u=(x-2)^0.5 then i solve for x in terms of u
so i get x= u^2+2
dx=2u du
[tex]\int\frac{\sqrt{x-2}} {x+1} \,dx=\int\frac{u}{u^2+3}\,2u\,du[/tex]
where when i substitute i get:
[tex]\int\frac{2u^2} {u^2+3}\,du[/tex]. after that I am completely lost.