Expert: Paul Klarreich - 10/11/2005

Question
How do you integrate the following function:        

     1/( sqrt(1+2x)+3 )

It is hard to do because the 3 is in the denominator but yet not in the square root.  Show steps if possible please.

Answer
Hi, Jon,
Try this scheme.  (If you get stuck after this, let me know and I'll send some more stuff.)

To integrate: Note my attempt at an integral symbol.

(
| [1/( sqrt(1+2x)+3 ) ] dx
)


Do the substitution  u = sqrt(1+2x),

BUT do it this way:  u^2 = 1 + 2x

Then:      2u du = 2dx, or  u du = dx

And the integral becomes:
[USE A FIXED FONT]

(  u du
| -------
) u + 3

which you handle by dividing out.  You will get:
 u           3
----- = 1 - -----
u + 3       u + 3

And these two terms are easily integrated.  Then you just put back your  u = sqrt(1 + 2x).

The answer will involve a logarithm and lots of roots and be a bit messy, but, er...  WELCOME TO CALCULUS II.