Expert: Paul Klarreich - 4/29/2007

Question
Hello
Im having some problem solving this integral... doesnt seems to have a simple solution.
Integral of sqrt(x^5/2 + x)dx
Which methos would be suitable?


Answer
Questioner:   white
Category:  Calculus
 
Subject:  Calculating an integral
Question:  Hello
Im having some problem solving this integral... doesnt seems to have a simple solution.
Integral of sqrt(x^5/2 + x)dx
Which methos would be suitable?
......................................
Hi, White,

This is just a simple substitution example (Har, har -- simple, he says, after it takes six pages to finish it.)

For your integral:

{
| sqrt(x^5/2 + x)dx
}

Start by factoring out an 'x' from inside the square root:
{
| sqrt( x( x^3/2 + 1) )dx =
}

{
| sqrt(x) sqrt(x^3/2 + 1)dx
}

Now do your 'simple' substition:

Let  u = sqrt(x^3/2 + 1) = sqrt(v),  where

v = x^3/2 + 1  << use chain rule to differentiate.
     
du       1
-- = ---------
dv   2 sqrt(v)

dv
-- = 3/2 x^1/2 = 3 sqrt(x) / 2
dx
         1            3 sqrt(x)
du = ----------------- ---------- dx
    2 sqrt(x^3/2 + 1)     2

       3 sqrt(x)
du = ----------------- dx
    4 sqrt(x^3/2 + 1)

Solve for dx, as in every 'simple substitution.'

    4 sqrt(x^3/2 + 1)
dx = ------------------- du
       3 sqrt(x)

        4u
dx = ---------- du
     3 sqrt(x)

Back to your integral:
{
| sqrt(x) sqrt(x^3/2 + 1)dx =
}

{                 4u
| sqrt(x) u  ----------- du =
}             3 sqrt(x)

Nice of those sqrt(x)'s to cancel.

{    4u
| u  --- du =
}     3

{ 4u^2
| ---- du =
}  3

4u^3/9

= (4/9) [sqrt(x^3/2 + 1)]^3

etc.

FULL DISCLOSURE STATEMENT.  There is a program called THE INTEGRATOR on the web, (look for Wolfram) which apparently can do anything.  Well, not really, and if it does it, you just get the answer.

But if you use it and get an answer, you can use that as a strong hint as to the method.