integration (x^2-x+6)/(x^3+3x) dx by parts

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summergrl

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integrate (x^2-x+6)/(x^3+3x) dx

I get stuck when im trying to find what a b and c should be.
 
I don't know what a, b or c are and it looks like you've tried to include an image; could you edit the post please so that I can help?
 
summergrl said:
integrate (x^2-x+6)/(x^3+3x) dx

I get stuck when im trying to find what a b and c should be.
Click to expand...
What "a", "b", and "c"? There is only "x" in your integrand.

Please reply with clarification, including a clear listing of all of your work and reasoning so far.

Thank you.

Eliz.
 
sorry...the image was there by mistake.
I am pretty sure I need to use integration by parts and we use A, B, and C.
So so far i seperated it into
A/X + (Bx+C)/(x^2+3)
 
I think you mean partial fractions, not integration by parts.

\(\displaystyle \L \frac{x^2 - x + 6}{x(x^2 + 3)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 3}\)

\(\displaystyle \L \frac{x^2 - x + 6}{x(x^2 + 3)} = \frac{A(x^2+3)}{x(x^2+3)} + \frac{x(Bx + C)}{x(x^2 + 3)}\)

\(\displaystyle \L x^2 - x + 6 = A(x^2 + 3) + x(Bx + C)\)

\(\displaystyle \L x^2 - x + 6 = (A + B)x^2 + Cx + 3A\)

equate coefficients ...

\(\displaystyle \L A + B = 1\)
\(\displaystyle \L C = -1\)
\(\displaystyle \L 3A = 6\)

can you finish?
 
\(\displaystyle \L x^2 - x + 6 = (A+B)x^2 + Cx + 3A\)

look at the left side of the equation ... what is the coefficient of x<sup>2</sup>?

look at the right side of the equation ... what is the coefficient of x<sup>2</sup>?

now set the coeffcients equal.
 
okay thanks! i understand that part now...so I got B=-1, C=-1, and A =2.
then i get the integration of
(2/x) - (x/x^2+3) - (1/x^2+3)
and i get 2lnx - 1/2ln(x^2+3) but then i don't know how to do the third part. Does it have anything to do with arctan?

p.s. whats this talk about pigs?? hehe
 
skeeter said:
Count Iblis said:
If you know about complex numbers you can do it much faster...
Click to expand...

... and if only pigs could fly.
Click to expand...

:D

But it is still true that many people who do know their complex analysis still do partial fractions in the tedious way. This is a modified version of my favorite way of doing partial fractions. After factoring the denominator we put:

\(\displaystyle \frac{x^2-x+6}{x(x^2+3)}=\frac{A}{x} + \frac{B + Cx}{x^2+3}\)

Multiply both sides by x and take the limit \(\displaystyle x\rightarrow 0\) to find A: The right hand side is then A and the left hand side is:

\(\displaystyle \lim_{x\rightarrow 0}x \frac{x^2-x+6}{x(x^2+3)} = \lim_{x\rightarrow 0} \frac{x^2-x+6}{x^2+3}=2\)

To find both B and C, multiply both sides by \(\displaystyle x^2 + 3\) and take the limit \(\displaystyle x\rightarrow i\sqrt{3}\) and equate real and imaginary parts. The right hand side then becomes:

\(\displaystyle B + i\sqrt{3}C\)

The left hand side is

\(\displaystyle \lim_{x\rightarrow i\sqrt{3}}(x^2+3) \frac{x^2-x+6}{x(x^2+3)} = \lim_{x\rightarrow i\sqrt{3}} \frac{x^2-x+6}{x}=-1-i\sqrt{3}\)
 
summergrl said:
okay thanks! i understand that part now...so I got B=-1, C=-1, and A =2.
then i get the integration of
(2/x) - (x/x^2+3) - (1/x^2+3)
and i get 2lnx - 1/2ln(x^2+3) but then i don't know how to do the third part. Does it have anything to do with arctan?

p.s. whats this talk about pigs?? hehe
Click to expand...

Use:

\(\displaystyle \int\frac{1}{x^2 + a^2}dx = \frac{1}{a}\arctan\left(\frac{x}{a}\right)\)
 
summergrl said:
okay thanks! i understand that part now...so I got B=-1, C=-1, and A =2.
then i get the integration of
(2/x) - (x/x^2+3) - (1/x^2+3)
and i get 2lnx - 1/2ln(x^2+3) but then i don't know how to do the third part. Does it have anything to do with arctan? yes ... 1/sqrt(3)*arctan[x/sqrt(3)]

p.s. whats this talk about pigs?? I just like pigs
Click to expand...
 
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