# Thread: any better or easier way to solve it ?

1. ## any better or easier way to solve it ?

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

2. you make a mistake, you only can use partial fractions when in the denominator is in factors, and in your problem has sums

3. ## sorry it was a mistake

yes ur right im sorry its a mistake .. it is a multiplication instead of addition in the denominator ..

is the way i showed in my prev. post the right way to approach it or there is a better and easier way to do it ?

thanx

4. $\int \frac{x^2+1}{(x^2+x+1)^2(x-2)^2}\,dx$

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

Solve the identity for $A, B, C, D, E, F$.

This is going to be one ugly integral.

5. ## it is really complicated

i got

(Ax+B)(x^3-3x^3+x^4+4)+(Cx+D)(x^2-4x+4)+E(x-2)(x^2+x+1)^2 +F(x^2+x+1)^2
=x^2+1

it got really bad i dont know how to go further
is there any better way or shortcuts ?
thnx

6. You can 'steal' the F by cross multiplying by $(x-2)^2$ and then letting x=2.
So, by inspection F=5/49.

7. ## how did u got the F

thank u

how did u do that for F?? i didnt get it

can we do the same things for the others or i have to expand everything

8. Originally Posted by alex83
thank u

how did u do that for F?? i didnt get it

can we do the same things for the others or i have to expand everything
$\frac{x^2+1}{(x^2+x+1)^2(x-2)^2} = \frac{Ax+B}{x^2+x+1}+\frac{Cx+D}{(x^2+x+1)^2}+\fra c{E}{x-2}+\frac{F}{(x-2)^2}$

Multiply both sides by $(x-2)^2$

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

And set x to 2

$\frac{2^2+1}{(2^2+2+1)^2} = F = \frac{5}{49}$

9. ## how to find other variables

thank u

but i've been playing around with it to figure out the rest A, B, C, D, and E
but couldnt reach anywhere
any help pls

10. There are no simple ways to obtain the remaining terms in this problem.
All techniques are more or less the same.
You can just multiply out and set the two polynominals equal to each other.
Then if you wish you can use matrices to solve for A...F.
Or one idea I never see people do, but it works.
Just let x=0,1,2,3,4, what ever it takes.
This too creates 6 equations and 6 unknows.
BUT you have no choice in this.
You can only steal the F.

11. You can also use complex numbers

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

Multiply both sides by $(x^2+x+1)^2$

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

And set x to j where $j=e^{\frac{2i\pi}{3}}$

$\frac{j^2+1}{(j-2)^2} = Cj+D$

Using $j^2=-j-1$ you can find C and D easily

For the other unknowns you can use
$\frac{x^2+1}{(x^2+x+1)^2(x-2)^2} = \frac{Ax+B}{x^2+x+1}+\frac{Cx+D}{(x^2+x+1)^2}+\fra c{E}{x-2}+\frac{F}{(x-2)^2}$

with x=0 then x=1 then x=-1
which give the 3 last equations

This is a little bit work but it can be done !

12. This one is tedious, no matter how you do it.
I used to do these and a lot worse ones in my classroom,
before we started using Maple at my school.
One reason I do like these computer programs.

13. ## is this right ??

Originally Posted by running-gag
You can also use complex numbers

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

Multiply both sides by $(x^2+x+1)^2$

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

And set x to j where $j=e^{\frac{2i\pi}{3}}$

$\frac{j^2+1}{(j-2)^2} = Cj+D$

Using $j^2=-j-1$ you can find C and D easily

For the other unknowns you can use
$\frac{x^2+1}{(x^2+x+1)^2(x-2)^2} = \frac{Ax+B}{x^2+x+1}+\frac{Cx+D}{(x^2+x+1)^2}+\fra c{E}{x-2}+\frac{F}{(x-2)^2}$

with x=0 then x=1 then x=-1
which give the 3 last equations

This is a little bit work but it can be done !
thanks a lot but i cant do it this way it seems more complex i tried this but dont know if its correct or no .

i expanded the whole equation and then organized it to this form

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

is that right or just a mess

thank u

14. Could you explain how you found E equal to 5/7 ?
This is not what I find ...