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

  • Thread starter Chadlee88
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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 dont know where to start with this question because of the 2x + 5 in the denomentor. Could someone please point me in the right direction.
 

Answers and Replies

  • #2
George Jones
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Could someone please point me in the right direction.
Complete the square.
 
  • #3
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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 plz help.

thanx again George
 
  • #4
George Jones
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In an appropriate manner, take the 4 outside the square root.
 
  • #5
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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 hav:

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.
 
  • #6
George Jones
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Factor the 4 out of both terms that are inside the square root.

Hint: 1 = 4 * 1/4.
 
  • #7
matt grime
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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.
 

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