The Student Room Group

Integration

Hi there.
I'm stuck in this qustion

ax+bx2+2x+2+cx+dx22x+2=1x4+4\frac{ax+b}{x^2+2x+2}+\frac{cx+d}{x^2-2x+2}=\frac{1}{x^4+4}

I'm having trouble working out the values of a,b,c,d.

I did:
(ax+b)(x^2-2x+2)+(cx+d)(x^2+2x+2)≡1
and got when x=0
(i) 2b+2d≡1
(ii)a=-c;
(iii)-2a+b+2c+d≡0;
(iv)2a-2b+2c+2d≡0

so c=-1/8; a=1/8
d=b=1/4

This doesn't seem right because i need to integrate the two fractions and it doesn't work.

Thanks.
Reply 1
Goldenratio
Hi there.
I'm stuck in this qustion

ax+bx2+2x+2+cx+dx22x+2=1x4+4\frac{ax+b}{x^2+2x+2}+\frac{cx+d}{x^2-2x+2}=\frac{1}{x^4+4}

I'm having trouble working out the values of a,b,c,d.

I did:
(ax+b)(x^2-2x+2)+(cx+d)(x^2+2x+2)≡1
and got when x=0
(i) 2b+2d≡1
(ii)a=-c;
(iii)-2a+b+2c+d≡0;
(iv)2a-2b+2c+2d≡0

so c=-1/8; a=1/8
d=b=1/4

This doesn't seem right because i need to integrate the two fractions and it doesn't work.

Thanks.


You can integrate the LHS whatever a,b,c,d are.

Make a sub of u = x+1 in the left fraction and v = x-1 in the right and it might become clearer.
Reply 2
Thanks!
Sorry I'm not seeing it.
Reply 3
Goldenratio
Thanks!
Sorry I'm not seeing it.


(ax+b)/(x^2+2x+2) becomes

au/(u^2+1) + (b-a)/(u^2+1)

which are standard integrals.
Reply 4
How did you split the fraction? Obviously there arent any factors so...
It's a STEP ii question, but i'm just not getting there. :p:
Thanks
Reply 5
Why couldn't I integrate the 1/(x^4+4) by substituting x^2=2tan^2p?
Reply 6
Goldenratio
How did you split the fraction? Obviously there arent any factors so...


What do you mean? They split the fraction for you :confused:

I just made the u = x+1 sub I'd previously suggested.
Reply 7
Okay thanks, got you!

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