- #1

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

I would be very greatfull. :D

I would be very greatfull. :D

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- Thread starter Penultimate
- Start date

- #1

- 26

- 0

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

I would be very greatfull. :D

I would be very greatfull. :D

- #2

HallsofIvy

Science Advisor

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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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- #4

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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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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:

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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