Integration

• Oct 13th 2009, 04:01 AM
mikegar813
Integration
integrate xSquareroot(x^2+2x+4) dx
• Oct 13th 2009, 04:33 AM
HallsofIvy
Quote:

Originally Posted by mikegar813
integrate xSquareroot(x^2+2x+4) dx

Complete the square: x^2+ 2x+ 4= x^2+ 2x+ 1+ 3= (x+1)^2+ 3. First let u= x+1, then use a trig substitution.

This is the third question you have asked with any indication of what you have tried yourself. If you do not show your own work, I will not respond to any more questions.
• Oct 13th 2009, 05:03 AM
mikegar813
i just needed help starting them i will work them and post my work and answers. thanks.
• Oct 13th 2009, 05:14 AM
mikegar813
if u=x+1
du=1 dx and when i substitute back in i get
integrate x Squareroot(u^2+3) dx
what do i do with the du since it doesn't match the x outside the squareroot?
• Oct 14th 2009, 03:55 AM
tonio
Quote:

Originally Posted by mikegar813
if u=x+1
du=1 dx and when i substitute back in i get
integrate x Squareroot(u^2+3) dx
what do i do with the du since it doesn't match the x outside the squareroot?

If you substitute u = x + 1 then x = u - 1. You cannot leave that x out there of the square root after you do this!

Tonio
• Oct 14th 2009, 10:22 AM
mikegar813
Quote:

Originally Posted by tonio
If you substitute u = x + 1 then x = u - 1. You cannot leave that x out there of the square root after you do this!

Tonio

The 'x' was already out side of the square root.
• Oct 14th 2009, 07:48 PM
tonio
Quote:

Originally Posted by mikegar813
The 'x' was already out side of the square root.

So what? When you make substitution in integrals you change ALL the x's for the substitution! You decided u = x + 1 <==> x = u-1 ? Then ALL the
x's in the integral are changed into u-1...!

Tonio