Expert: Ahmed Salami - 2/8/2006

Question
Integrate ((t-1)^1/2)/(t-2) by substituting

u = (t-1)^1/2

Answer
Hi Farooq,
Sorry for the time it took.
To find the integral of [sqrt(t-1)]/t-2 by the
substitution u = sqrt(t-1), we have
u = (t-1)^(1/2)
du/dt = (1/2)(t-1)^(1/2 - 1)
     = (1/2)(t-1)^(-1/2)
     = (1/2)/(t-1)^(1/2)
     = (1/2)/u = 1/2u
dt = 2u du
Also,
u^2 = t - 1
t = u^2 + 1  i.e
t - 2 = (u^2 + 1) - 2 = u^2 - 1

The integral $[sqrt(t-1)]/(t-2) dt becomes
($ represents the integral sign)
$u/(u^2 - 1) 2u du
= $2u^2/(u^2 - 1)  du = 2$u^2/(u^2 - 1) du
A little trick would come in handy here
2$u^2/(u^2 - 1) du = 2$(u^2 - 1 + 1)/(u^2 - 1) du
= 2$[(u^2 - 1)/(u^2 - 1)  + 1/(u^2 - 1)] du
= 2$[1 + 1/(u^2 - 1)] du
By partial fractions,
1/(u^2 - 1) = (1/2)[1/(u-1) - 1/(u+1)]

2$[1 + 1/(u^2 - 1)] du becomes
2$1 + (1/2)[1/(u-1) - 1/(u+1)]  du
= $[2 + 1/(u-1) - 1/(u+1)] du
= 2u + ln(u-1) - ln(u+1) + c   (c is constant)
= 2u + ln[(u-1)/(u+1)] + c
= 2sqrt(t-1) + ln[sqrt(t-1) - 1 / sqrt(t-1) + 1] + c

I hope i have helped. You can always get back to me.
Regards.