Solution to Problem 11.8-4


© 2001 by Karl Hahn 		KCT logo

4)The problem is to integrate 


   


    dx
    _____
 x √x - a

Step 1: Substitute to make a single variable appear under the radical. This will whip this thing into shape where we can apply the square root substitution. Work that through and then click here to continue.

Your substitution is simply 


   u  =  x - a
hence 

u + a  =  x
and 

du  =  dx

so the integral becomes 


   


    dx
    _____  =
 x √x - a


  


     du
          _
 (u + a) √u

Step 2: Make the square root substitution. This is the one you learned about in this section. When you've done it, click here to continue.

The substitution is 


   v2  =  u
and 

2v dv  =  du

So the integral becomes 


   


    dx
    _____  =  2
 x √x - a


 


    v dv
             =  2
 (v2 + a) v


 


    dv
         
 (v2 + a)

Step 3: Divide out the constant. Until now I've given you numbers for constants, but here we just have the symbol, a, as a constant. But the procedure is no different. Work it through and then click here to continue.

You should have 


   


    dx
    _____  =  2
 x √x - a


 


    dv
           =
 (v2 + a)


  2
   
  a


 


      dv
     _      
 (v/√a)2 + 1

Step 4: Apply trig substitution. We've done integrands in this form a bunch of times now, so this shouldn't be a problem for you. When you've got it, click here to continue.

The trig substitution is 


               v
   tan(w)  =   _
              √a
and 

 _
√a (tan2(w) + 1) dw  =  dv

The integral becomes 


   


    dx
    _____  =
 x √x - a


   2
   _
  √a


 


 tan2(w) + 1
             dw  =
 tan2(2) + 1


   2
   _
  √a


 


        2w
 dw  =   _  +  C
        √a

The actual integration at this point is so trivial that I didn't break it out to a separate step. So the next step is

Step 5: Substitute back. You have to go from w to v to u to x. Work backward through your substitutions and then click here to see the final answer.

The back substitution is progressively 


   


    dx
    _____  =
 x √x - a


  2w
   _  +  C  =
  √a


   2
   _ arctan
  √a





 v
 _
√a






  +  C  =



             2
             _ arctan
            √a




 _
√u
 _
√a






  +  C  =



   2
   _ arctan
  √a




 _____
√x - a
   _  
  √a






  +  C

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