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Homework Help: Integral Solving?

  1. Jun 18, 2009 #1

    I would be very greatfull. :D
  2. jcsd
  3. Jun 18, 2009 #2


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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]
  4. Jun 18, 2009 #3


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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.
  5. Jun 18, 2009 #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.
  6. Jun 18, 2009 #5


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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
  7. Jun 19, 2009 #6


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    You check your answers by differentiating them.
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