1. ## Integration

integrate xSquareroot(x^2+2x+4) dx

2. 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.

3. i just needed help starting them i will work them and post my work and answers. thanks.

4. 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?

5. 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

6. 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.

7. 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