Turn on thread page Beta
    • Thread Starter
    Offline

    0
    ReputationRep:
    Can anyone help me integrate this:

    (x+2)/(x²+6x+4)^½

    the answer is [(x+3)²-5]^½ - arcosh[(x+3)/5^½] + c

    I can see where the arcosh comes from with completing the square and all but I don't know how you go about getting there. Do you split it up and do it by parts? Help me!
    Offline

    1
    ReputationRep:
    Let u = x^2 + 6x + 4
    du/dx = 2x + 6

    1/2.du - 1 = x+2 which is the numerator, so

    The integral becomes
    Int (1/(2u^(1/2)) du - Int 1/((x+3)^2 - 5) dx

    I think.
    • Thread Starter
    Offline

    0
    ReputationRep:
    I'll try it right now! Thanks
    Offline

    0
    ReputationRep:
    another method:
    write the integral as:
    (x+3-1)/(x²+6x+4)^½
    which is ∫(x+3)/(x²+6x+4)^½ plus ∫-1/(x²+6x+4)^½
    the second one u can do via the completing square and arcosh formula
    in the first one, 2x+6 is the deravative of the bracket in the bottem, so try
    differentiating (x²+6x+4)^½, and u'll get the integral.
 
 
 

1,086

students online now

800,000+

Exam discussions

Find your exam discussion here

Poll
Should universities take a stronger line on drugs?

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

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.