How Do You Integrate dx/sqrt(x^2+2x+5)?

[Chadlee88]

Homework Statement

Integrate: dx/sqrt(x^2+2x +5)

Homework Equations

refer to above

The Attempt at a Solution

I can integrate the equation dx/sqrt(1+x^2) using the rules

cosh^2 u - sinh^2 u = 1
cosh^2 u = 1+sinh^2 u

but i don't know where to start with this question because of the 2x + 5 in the denomentor. Could someone please point me in the right direction.

 

[George Jones]

Could someone please point me in the right direction.

Complete the square.

 

[Chadlee88]

Complete the square.

Thanks that helped but I'm still stuck with a 4 i don't know to get rid of :S

integral of: dx/sqrt(x^2+2x+5)

Equals the integral of: dx/sqrt((x+1)^2 + 4)

Using:
1. (x+1) = sinh u
2. cosh^2 u = 1 + sinh^2 u
3. dx = cosh u du

I get to this stage:

Integral of: dx/sqrt((x+1)^2 + 4)

Equals the integral of: cosh u du/((x+1)^2 + 4)

This is where i get stuck, I'm not sure what to do with the 4. could som1 please help.

thanx again George

 

[George Jones]

In an appropriate manner, take the 4 outside the square root.

 

[Chadlee88]

In an appropriate manner, take the 4 outside the square root.

ya i squared everything but the bottom line is not in the correct form.

i have:

cosh ^2 u du / (x+1)^2 +4

if the denomentator was (x+1)^2 + 1 i could just substitute but cos of that 4 i can't. gettin rid of taht 4 is my problem.

 

[George Jones]

Factor the 4 out of both terms that are inside the square root.

Hint: 1 = 4 * 1/4.

 

[matt grime]

if the denomentator was (x+1)^2 + 1 i could just substitute but cos of that 4 i can't. gettin rid of taht 4 is my problem.

There is more than one possible substitution you can make, you know.