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

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\int\frac{x+5}{2x^2+2x+3}

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 2x^2+ 2x+ 3= 0 are
\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 /> 2x^2+ 2x+ 3= 2(x^2+ x)+ 3<br /> 2(x^2+ x+ 1/4- 1/4)+ 3= 2(x^2+ x+ 1/4)- 1/2+ 3[/itex]&lt;br /&gt; = 2(x+ 1/2)^2+ 5/2[/itex]&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; Try the substitution u= x+ 1/2 and remember that &amp;lt;br /&amp;gt; \int\frac{dx}{x^2+ 1}= arctan(x)+ C
 
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.
 
\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}

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

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

Make the substitution w=sqrt(4/5u^2)=u*sqrt(4/5), use \int\frac{dx}{x^2+1}=arctan(x), 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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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