Solve Integral: $\int\frac{x+5}{2x^2+2x+3}$

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Homework Help Overview

The discussion revolves around the integral $\int\frac{x+5}{2x^2+2x+3}$, which falls under the subject area of calculus, specifically integration techniques.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss factoring the denominator and using partial fractions, but note that the quadratic has no real roots. There are suggestions to complete the square and to consider substitutions for integration. Some participants express a lack of time to engage deeply with the problem and inquire about online integral solvers.

Discussion Status

The discussion includes various approaches to the integral, with some participants providing potential methods and substitutions. There is a recognition of the need for further exploration of the integral, and while some guidance has been offered, there is no explicit consensus on a complete solution.

Contextual Notes

One participant mentions an upcoming exam and limited time to work on the problem, indicating a constraint that may affect their engagement with the discussion.

Penultimate
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[tex]\int[/tex][tex]\frac{x+5}{2x^2+2x+3}[/tex]

I would be very greatfull. :D
 
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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]<br /> This has no real roots so certainly cannot be factored with real coefficients.<br /> <br /> So complete the square in the denominator:<br /> [tex]2x^2+ 2x+ 3= 2(x^2+ x)+ 3[/tex]<br /> [tex]2(x^2+ x+ 1/4- 1/4)+ 3= 2(x^2+ x+ 1/4)- 1/2+ 3[/itex]<br /> [tex]= 2(x+ 1/2)^2+ 5/2[/itex]<br /> <br /> Try the substitution u= x+ 1/2 and remember that <br /> [tex]\int\frac{dx}{x^2+ 1}= arctan(x)+ C[/tex][/tex][/tex][/tex]
 
Would be nice if you actually bothered to attempt it and show us your attempts. You could try completing the square.
 
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.
 
[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
 
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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