The Student Room Group

Reply 1

u^2 = x-1
2u.du/dx = 1
dx = 2u.du

x=2, u=1
x=5, u=2

=> INT (2u/1+u)du , from 1 to 2

= 2 INT [(1+u)-1 / (1+u)] du

= 2 INT [1 - 1/(1+u)] du

= 2[u - ln(1+u)] , from 1 to 2

= 2[(2-ln3) - (1-ln2)]

= 2[1 + ln(2/3)]

= 2 + 2ln(2/3)

Reply 2

u^2=x-1 --> dx/du=2u
Plugging in, you want to integrate
2u/(1+u) with respect to u
split this into u/(1+u) + u/(1+u)
These integrate into
(u+1) - ln(u+1) + (u+1) - ln(u+1)=2(u+1)-2ln(u+1)
since we want to integrate from x=2 to x=5, we want u from 1 to 2
plugging in, we get 2(3)-2ln(3)-2(2)+2ln(2)= 2 + 2(ln(2/3))=1.19

Reply 3

mockel
u^2 = x-1
2u.du/dx = 1
dx = 2u.du

x=2, u=1
x=5, u=2

=> INT (2u/1+u)du , from 1 to 2

= 2 INT [(1+u)-1 / (1+u)] du

= 2 INT [1 - 1/(1+u)] du


= 2[u - ln(1+u)] , from 1 to 2

= 2[(2-ln3) - (1-ln2)]

= 2[1 + ln(2/3)]

= 2 + 2ln(2/3)


Thank you to both of you.

I have highlighted the step that I couldn't do. More practice needed I guess.