any easier way to integrate (x^2+1)/[(x^2+x+1)^2+(x-2)^2] ?

[alex83]

integral (x^2+1)/[(x^2+x+1)^2 + (x-2)^2]

what i tried is :

A/(x-2) + B/(x-2)^2 + (Cx+D)/(x^2+x+1) + (Ex+F)/(x^2+x+1)^2

then try to find ABCDEF

its seems too long and complicated is there any better way to solve it pls

thank u

 

[galactus]

Re: any easier way to integrate this pls.

$\displaystyle \int \frac{x^{2}+1}{(x^{2}+x+1)^{2}+(x-2)^{2}}dx$

Are you sure that is it?. This is horrendous.

 

[alex83]

Re: any easier way to integrate this pls.

$\displaystyle \int \frac{x^{2}+1}{(x^{2}+x+1)^{2}+(x-2)^{2}}dx$

Are you sure that is it?. This is horrendous.

yes that is it ??

is the way i started with the only one that can b used ?? is it right ??

thanks

 

[galactus]

No, I am sorry to say. The addition in the denominator is not a partial fraction. It is more complicated than that.

 

[galactus]

From where are you getting these monster integrals?. I ran this one through Maple just to see what it would give me and it was a mess.

I would not try it by hand. This one has no nice closed indefinite form.

 

[alex83]

i just emailed my prof and then he replied to all the class to appologize that it should be a multiplication instead of addition in the denominator

so whats the best way to approach it from there

thank u and sorry for the confusion

 

[galactus]

OK. That makes a lot of difference. Indeed, partial fractions would be a way to go.

$\displaystyle \frac{Ax+B}{(x^{2}+x+1)^{2}}+\frac{Cx+D}{x^{2}+x+1}+\frac{E}{(x-2)^{2}}+\frac{F}{x-2}=\frac{x^{2}+1}{(x^2+x+1)^2 (x-2)^2}$

Your approach is correct. Now to lumber through the algebra. That is the worst part of these.

 

[alex83]

cant get anywhere with the problem any help on how to set the (matrix) or equations for A,B,C,D,E.F

thank u

 

[stapel]

cant get anywhere with the problem any help on how to set the (matrix) or equations for A,B,C,D,E.F

Do the multiplication, like you learned back in algebra, for getting rid of the denominators in this rational equation. You should get:

. . . . .$\displaystyle (Ax\, +\, B)(x\, -\, 2)^2\, +\, (Cx\, +\, D)(x^2\, +\, x\, +\, 1)(x\, -\, 2)^2$
. . . . . . . . .$\displaystyle +\, E(x^2\, +\, x\, +\, 1)\, +\, F(x\, -\, 2)(x^2\, +\, x\, +\, 1)^2\, =\, x^2\, +\, 1$

Pick values for x; plug them in, and simplify to create equations in A, B, C, D, E, and F. For instance, if x = 2, then:

. . . . .$\displaystyle E(4\, +\, 2\, +\, 1)\, =\, 4\, +\, 1$

...so E = 5/7. If x = 0, then:

. . . . .$\displaystyle (B)(-2)^2\, +\, (D)(1)(-2)^2\, +\, \frac{5}{7}(1)\, +\, F(-2)(1)^2\, =\, 1$

. . . . .$\displaystyle 4B\, +\, 4D\, +\, \frac{5}{7}\, -\, 2F\, =\, 1$

...and so forth. Once you have four more equations, set up your matrix and solve.

If you get stuck, please reply showing your steps and reasoning. Thank you!

 

[alex83]

Re:

[quote If you get stuck, please reply showing your steps and reasoning. Thank you!

first of all thanks a lot

what i did is that i expanded the whole equation (which was a mess :x ) and got the following:

A(x^5-3x^4+x^3+4x) + B(x^4-3x^3+x^2+4) + C(x^3-4x^2+4x) +D(x^2-4x+4) +E(x^5-x^3-4x^2+5x-2) +F(x^4+2x^3+3x^2+2x+1) = x^2 + 1
from this i got the following

A+E =0 (for x^5)
-3A+B+F =0 (for x^4)
A-3B+C-E+2F =0 (for x^3)
B-4C+D-4E+3F =1 (for x^2)
4A+4C-4D+5E+2F=0 (for x)
4B+4D-2E+F =1 (for cnst)

ok i already got E as u explained (E=5/7)
and F=(5/49)

and started to substitute
in the equations above so i got
A=-E= -5/7
B= -35/49

AM I DOING THAT RIGHT ??? R THESE RESULTS CORRECT SO I CAN KEEP GOING OR THERE'S A MISTAKE SOMEWHERE ??

thanks again and again