Solution to Problem 11.8-5


© 2001 by Karl Hahn 		KCT logo

5)The problem is to integrate 


   


      dx
     _______
 ex √e2x + 1

Step 1: Use the exponential substitution that you learned in this section. Remember that  e2x  is nothing more than the square of  ex.  When you've got this one substituted, click here to continue.

The substitution is 



   u  =  ex
and 

du
    =  dx
 u

So the substituted integral is 


   


      dx
     _______  =
 ex √e2x + 1


  


     du
     ______
 u2 √u2 + 1

Step 2: Apply trig substitution. Click here if you need a reminder on what trig substitution to make here. This problem is now very similar to worked example number 3. You can do it by trig substitution or by hyperbolic substitution. Here I will do it using the former, but if you are motivated you can work it through again using hyperbolic substitution on your own. Make the substitution and click here to continue.

The substitution is 


   tan(v)  =  u
and 

(tan2(v) + 1) dv  =  du

The integral becomes 


   


      dx
     _______  =
 ex √e2x + 1


  


   (tan2(v) + 1) dv
          ___________  =
 tan2(v) √tan2(v) + 1


  

  ___________
 √tan2(v) + 1
              dv
    tan2(v)

Step 3: Use trig identities to simplify the whoopie out of this integrand. It is useful to get this one into terms of sines and cosines. Use identities from our recent table. Also remember that 



   tan2(v)  =



  sin2(v)
         
  cos2(v)
and 


sec(v)  =



     1
        
  cos(v)

Make the simplifications and then click here to continue.

The simplification steps are, first in the numerator: 

    ___________
   √tan2(v) + 1  =  sec(v)

Make that replacement, then convert the whole thing to sines and cosines. You find that 


  ___________
 √tan2(v) + 1
               =
    tan2(v)



       1
          
    cos(v)
               =
    sin2(v)
           
    cos2(v)

   cos(v)
         
  sin2(v)

So now the integral is 


   


      dx
     _______  =
 ex √e2x + 1


  


  cos(v)
         dv
 sin2(v)

Step 4: Simple substitution. Do you see the one part of this integrand that is the derivative of the other? Make the substitution suggested by that and then click here to continue.

The substitution is 


   w  =  sin(v)
and 

dw  =  cos(v) dv

When you make that substitution, the integral becomes trivial. 


   


      dx
     _______  =
 ex √e2x + 1


  


  cos(v)
         dv  =
 sin2(v)


  


 dw
     =
 w2


    1
  -    +  C
    w

Step 5: Substitute back. You have to go from w to v to u to x. You might have to use the identity table to do the back substitution of the trig functions. When you've done all of this, click here to see the final answer.

The progressive back substitutions are 


   


      dx
     _______  =
 ex √e2x + 1


    1                1
  -    +  C  =  -         +  C  =
    w             sin(v)

      ___________
     √tan2(v) + 1
   -               +  C  =
        tan(v)

     ______
    √u2 + 1
  -          +  C  =
       u

     _______
    √e2x + 1
  -           +  C
       ex

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