Partial Fractions Q8 (1 Viewer)

[want2beSMART]

INTEGRAL of (x^3+2)/x(x^2+1)dx

 

[Slidey]

x^3+2=(Ax+B)x+C(x^2+1)
sub x=-1, 0, 1 to get:
1=A-B+2C
2=C
3=A+B+2C

-3=A-B
-1=A+B
A=-2, B=1
Integral of (-2x+1)/(x^2+1) + 2/x = - 2x/(x^2+1) + 1/(1+x^2) + 2/x
= -ln(x^2+1)+2lnx+tan^-1x + C
= ln[x^2/(x^2+1)] + tan^-1x + C

I think. I haven't covered partial fractions or inverse trig, yet... or integrals of things like 1/x, for that matter.

 

[want2beSMART]

erm the answer is x+2lnx - ln (x^2 + 1) - tan-1 x + C

 

[shafqat]

i think substitution of -1 is wrong slide rule

 

[want2beSMART]

shafqat do you have the answer?

 

[Euler]

Degrees are the same.

do something before you use partial fractions.

long division is one way to go.

 

[Slidey]

(x^3+2)/(x^3+x)=(x^3+x)/(x^3+x) + (2-x)/(x^3+x)
=1-(x-2)/(x^3+x)

(x-2) = (Ax+B)x + C(x^2+1)
Sub in x=-1,0,1:
-3=A-B+2C
-2=C
-1=A+B+2C

1=A-B
3=A+B
A=2, B=1, C=-2

So the integral of (x^3+2)/(x^3+x) is
Integral of [1-(2x+1)/(x^2+1)+2/x}]
= x - ln(x^2+1) - tan^-1x + 2lnx + C

BTW: Why was my initial method incorrect? All of my substitutions were correct, as far as I can see.

 

[KFunk]

Like Euler suggested, the partial fractions technique only works on P(x)/Q(x) where deg P < deg Q

I think it has to do with the fundamental theorem of algebra and the fact that if two polynomials of degree n are equal for more than n values of x, then they are equal for all values of x... which can be used to prove that the partial fractions you've split the polynomial into are identically equal to what's on the other side. The polynomials on either side of the equation that you created in your first post weren't of the same degree hence the partial fractions technique can't be used. You'll have to look into a proof of the fundamental theorem yourself if you want to know exactly why this is the case.

 

[lfc_reds2003]

you can use long division in that one

cos the degree of numerator is equal to degree of denominator

 

[want2beSMART]

can you tell me how to do long division?

 

[KFunk]

Martin gave a pretty good explanation here:

http://www.boredofstudies.org/community/showthread.php?t=60249

 

[want2beSMART]

haha dw i know how to do long division

but dont know when to use it...

when the power in the numerator is greater than the power in the denominator?

 

[lfc_reds2003]

greater than or equal to n then u sud have either logs inverse trigs or must use partials....