this from a calc text:
integral of sqrt(x-1)/x
wolfram spits it out but gives no indication as to what substitution was made

Question

anonymous · Answer

$$\int (x-1)^{0.5}x^{-1} dx$$ let u=sqrt(x-1)

anonymous · Answer

$$u=\sqrt(x-1), x=u^2+1, dx=2\sqrt{x-1}du, dx=2u$$
$$\int (x−1)^{0.5}x^{−1}dx= \int\frac{u}{u^2+1} 2udu$$$$= 2\int \frac{u^2}{u^2+1} du$$$$=2\int1-\frac{1}{u^2+1}du$$$$=2u-2tan^{-1}{u}+c$$$$=2\sqrt{x-1}-2tan^{-1}{\sqrt{x-1}}+c$$

anonymous · Answer

thank you very much Zed. i was going along that very same path to solution-i failed to make that algebraic operation where you broke the integrand into 2 parts

myininaya · Answer

u^2=u^2+1-1 :)

anonymous · Answer

yeah long divisions a pain :) here's a website that has the inverse trig integrals http://www.math.com/tables/integrals/tableof.htm

anonymous · Answer

ok

myininaya · Answer

sometimes doing the long division is easier though

myininaya · Answer

but definitely not in this case
gj by the way zed

anonymous · Answer

ok, so you divided u^2 by (u^2+1) to change the  expression to 2 parts. now I see thanx again