You are Here: Home

# Quick P3 Integration using Substitution qu. watch

Announcements
1. Question.. "using u^2 = x - 1, find ∫(1 / (1 + √(x-1))) .dx"

so 2u . du = 1
du = 1/(2u)

therefore ∫(1 / (1 + √(x-1))).dx = ∫(1 / (1 + u)) .2u .du

∫(2u / (1 + u)) .du = ??! looks so easy i seem to get lost!!

Cheers, SAHIR
2. (Original post by Sahir)
Question.. "using u^2 = x - 1, find ∫(1 / (1 + √(x-1))) .dx"

so 2u . du = 1
du = 1/(2u)

therefore ∫(1 / (1 + √(x-1))).dx = ∫(1 / (1 + u)) .2u .du

∫(2u / (1 + u)) .du = ??! looks so easy i seem to get lost!!

Cheers, SAHIR
∫(u / (1 + u)) .du = ∫ ((u+1) - 1) /(1+u) = ∫ 1 - (1/(1+u)) du = u - ln|u+1|.
3. 2u.du = 1.dx, so you replace the dx with 2u.du, not 1/2u .du
4. duuuuh.. i feel so stooopid!

Heh, cheers theone... rapid response maths at its best.
5. (Original post by elpaw)
2u.du = 1.dx, so you replace the dx with 2u.du, not 1/2u .du
Yeah that was just a lapse in concentration when i typed it.. wrote down it correctly and its correct in the next line.. cheers though elpaw.
6. (Original post by theone)
∫(u / (1 + u)) .du = ∫ ((u+1) - 1) /(1+u) = ∫ 1 - (1/(1+u)) du = u - ln|u+1|.
+ C!

and then you have to substitute x back in.
7. (Original post by elpaw)
+ C!

and then you have to substitute x back in.
Always there to pick me up when I fall

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: March 2, 2004
Today on TSR

### Exam Jam 2018

Join thousands of students this half term

### Solo - A Star Wars Story

Poll

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE