Question

Integral sqrt(x-2)/sqrt(x-1)

Answer 1

try u=sqrt(u-1)
i will be right back

Answer 2

\[u^2=x-1 => 2u u'=1 => u'=\frac{1}{2 u} => \frac{du}{dx}=\frac{1}{2 \sqrt{x-1}}\]
\[2 du=\frac{dx}{\sqrt{x-1}}\]
\[\int\limits_{}^{} \sqrt{u^2-1} \cdot 2 du=2 \int\limits_{}^{}\sqrt{u^2-1} du\]
since sqrt(x-2)=sqrt(x-1-1)=sqrt(u^2-1)
this part looks obvious to be to use a trig sub now

Answer 3

Thanks a bunch!

Answer 4

:)