Expert: Paul Klarreich - 1/10/2010

Question
How would i go about integrating (x^2+1)^1/2 by parts?

Answer
Questioner:   Tom
Category:  Calculus
Private:  No
 
Subject:  Integration
Question:  How would i go about integrating (x^2+1)^1/2 by parts?

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Give your integral a name:  (John, Shirley, whatever)

Call it K.

K = INT (x^2+1)^1/2  dx


u = (x^2 + 1)^1/2,   du = x/(x^2 + 1)^1/2 dx

dv = dx,  v = x

uv = x(x^2 + 1)^1/2

-INT v du = - INT x^2/(x^2 + 1)^1/2 dx


So K = x(x^2 + 1)^1/2 - INT x^2/(x^2 + 1)^1/2 dx

Now you just have to integrate that part.
           x^2
= - INT --------------- dx
       (x^2 + 1)^1/2

I would try this:

        x^2 + 1 - 1
= - INT -------------- dx
       (x^2 + 1)^1/2


                               1
= -INT [ (x^2 + 1)^1/2 -  -------------- ]dx
                          (x^2 + 1)^1/2

                               1
= INT [- (x^2 + 1)^1/2 +  ------------- ]dx
                         (x^2 + 1)^1/2


                    1
= - K + INT [  ------------- ]dx
              (x^2 + 1)^1/2


When you bring the K over, you have 2K on the left, and a uv and one more integral to do.

To do that last one, use a trig substitution.

Let  x = tan t,  dx = sec^2 t dt

             1
INT [  ------------- ]dx
      (x^2 + 1)^1/2

                  1
=   INT [  ----------------- ]sec^2 t dt
          (tan^2 t + 1)^1/2

                  1
=   INT [  ----------------- ]sec^2 t dt
          (sec^2 t)^1/2

              1
=   INT [  -------- ]sec^2 t dt
           sec t


=   INT sec t dt

This is a standard 'trick' integral (no, I don't mean 'trig' integral) that you can look up.  You can finish up from here.