Integral Solving?

  • #1
[tex]\int[/tex][tex]\frac{x+5}{2x^2+2x+3}[/tex]

I would be very greatfull. :D
 

Answers and Replies

  • #2
HallsofIvy
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The first thing to do is to try to factor the denominator, then use "partial fractions".
Using the quadratic formula, the roots of [itex]2x^2+ 2x+ 3= 0[/itex] are
[tex]\frac{-2\pm\sqrt{2^2- 4(2)(3)}}{4}= \frac{-2\pm\sqrt{-24}}{5}[/itex]
This has no real roots so certainly cannot be factored with real coefficients.

So complete the square in the denominator:
[tex]2x^2+ 2x+ 3= 2(x^2+ x)+ 3[/tex]
[tex]2(x^2+ x+ 1/4- 1/4)+ 3= 2(x^2+ x+ 1/4)- 1/2+ 3[/itex]
[tex]= 2(x+ 1/2)^2+ 5/2[/itex]

Try the substitution u= x+ 1/2 and remember that
[tex]\int\frac{dx}{x^2+ 1}= arctan(x)+ C[/tex]
 
  • #3
Cyosis
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Would be nice if you actually bothered to attempt it and show us your attempts. You could try completing the square.
 
  • #4
Thanks for the answears.
I haven`t got any time at all , tomorow i have the exam and i am studying some other exercises.
I haven`t done much efforts on this one but i am asking you if you have got some time to solve it.I know it sounds strange but i still hope somebody with a god will would help.
Another thing i wanted to ask : is there any integral solver online.(i want that to confirm the integrals i have solved). Thank you again.
 
  • #5
ideasrule
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[tex]\int\frac{(u+9/2)du}{2u^2+5/2}= \int\frac{u du}{2u^2+5/2}+\int\frac{9/2 du}{2u^2+5/2}[/tex]

The first integral should be easy; just make the substitution w=2u^2+5/2. For the second integral:

[tex]\int\frac{9/2 du}{2u^2+5/2}=\int\frac{9/5du}{4/5u^2+1}[/tex]

Make the substitution w=sqrt(4/5u^2)=u*sqrt(4/5), use [tex]\int\frac{dx}{x^2+1}=arctan(x)[/tex], and everything else should be easy.

As for an online integrator, try http://integrals.wolfram.com/index.jsp?expr=(x%2B5)%2F(2x^2%2B2x%2B3)&random=false
 
  • #6
Cyosis
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Penultimate said:
Thanks for the answears.
I haven`t got any time at all , tomorow i have the exam and i am studying some other exercises.
I haven`t done much efforts on this one but i am asking you if you have got some time to solve it.I know it sounds strange but i still hope somebody with a god will would help.
Another thing i wanted to ask : is there any integral solver online.(i want that to confirm the integrals i have solved). Thank you again.

You check your answers by differentiating them.
 

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