1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Find the indefinite integral by u-sub

  1. Apr 20, 2013 #1
    1. The problem statement, all variables and given/known data

    [itex]\int1/(1+\sqrt{2x})\,dx[/itex]



    2. Relevant equations

    [itex]u=1+\sqrt{2x} \Rightarrow \sqrt{2x}=u-1[/itex]
    [itex]du=1/\sqrt{2x}dx \Rightarrow \sqrt{2x}du=dx[/itex]


    3. The attempt at a solution

    [itex]\int1/(1+\sqrt{2x})\,dx = \int\sqrt{2x}/(1+\sqrt{2x})\,du = \int(u-1)/u\,du = \int\,du-\int1/u\,du = u-ln|u|+C = 1+\sqrt{2x}-ln|1+\sqrt{2x}|+C[/itex]

    The book im using has the answer as:
    [itex]\sqrt{2x}-ln|1+\sqrt{2x}|+C[/itex]

    Where am i going wrong?
     
    Last edited: Apr 20, 2013
  2. jcsd
  3. Apr 20, 2013 #2

    Curious3141

    User Avatar
    Homework Helper

    That "extra" 1 in your answer can be "absorbed" into the arbitrary constant C (so you can drop the 1). The answers are equivalent.
     
  4. Apr 20, 2013 #3
    Oh ok. Thank you curious.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Find the indefinite integral by u-sub
  1. U-sub trig integrals (Replies: 2)

Loading...