# Integral Solving?

1. Jun 18, 2009

### Penultimate

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

I would be very greatfull. :D

2. Jun 18, 2009

### HallsofIvy

Staff Emeritus
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
$$\frac{-2\pm\sqrt{2^2- 4(2)(3)}}{4}= \frac{-2\pm\sqrt{-24}}{5}[/itex] This has no real roots so certainly cannot be factored with real coefficients. So complete the square in the denominator: [tex]2x^2+ 2x+ 3= 2(x^2+ x)+ 3$$
$$2(x^2+ x+ 1/4- 1/4)+ 3= 2(x^2+ x+ 1/4)- 1/2+ 3[/itex] [tex]= 2(x+ 1/2)^2+ 5/2[/itex] Try the substitution u= x+ 1/2 and remember that [tex]\int\frac{dx}{x^2+ 1}= arctan(x)+ C$$

3. Jun 18, 2009

### Cyosis

Would be nice if you actually bothered to attempt it and show us your attempts. You could try completing the square.

4. Jun 18, 2009

### Penultimate

Thanks for the answears.
I havent got any time at all , tomorow i have the exam and i am studying some other exercises.
I havent 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. Jun 18, 2009

### ideasrule

$$\int\frac{(u+9/2)du}{2u^2+5/2}= \int\frac{u du}{2u^2+5/2}+\int\frac{9/2 du}{2u^2+5/2}$$

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

6. Jun 19, 2009