Partial Fractions Integral

Printable View

Nov 2nd 2008, 10:32 PM
U-God

Partial Fractions Integral

I have a feeling that I made my mistake in this question near the end, but I'll write it all out anyway:

$\int \frac{4x+3}{(x^2+1)(x^2+2)}dx$

$\frac{4x+3}{(x^2+1)(x^2+2)} = \frac{Ax+B}{x^2+1} + \frac{Cx+D}{x^2+2}$

$(Ax + B)(x^2 + 2) + (Cx+D)(x^2 +1) = 4x + 3$

$4x + 3 = Ax^3 + Cx^3 + Bx^2 + Dx^2 + 2Ax + Cx + 2B + D$

equated co-efficients to get :

$4\int\frac{x}{x^2 + 1}dx + 3\int\frac{1}{x^2 + 2}dx - 4\int\frac{x}{x^2 + 2}dx - 3\int\frac{1}{x^2 + 2}dx$

I can do the second and fourth integrals fine, so I'll just show work for the first and third.

First $4\int\frac{x}{x^2 + 1}dx$ let $x = tan(u)$

therefore $dx = sec^(u)du$

now I sub them in: $4\int\frac{tan(u)}{sec^2(u)}sec^2(u)du$

$4\int\ tan(u)du = -4log[cos(u)] + E$

from the substitution I derived: $cos(u) = \frac{1}{\sqrt{1 + x^2}}$

therefore $= 3log(1 + x^2) + E$

I'll stop here, because I think that by here I've made a mistake (this part of my answer does not match the correct answer). If anyone can spot it, it would be great.
Thanks,
Forgive typos, this one was quite tedious for me to write out.
Nov 2nd 2008, 10:45 PM
badgerigar

Everything here looks fine, except for the last line of your working:

Quote:

therefore http://www.mathhelpforum.com/math-he...acc353a4-1.gif

I don't know what is intended to be on the left of that = but I am guessing that something went wrong there.

Everything you have posted is entirely correct and will lead to the correct solution, but I think it is worth pointing out that a substitution $u = x^2+1$ would have been easier.
Nov 2nd 2008, 10:49 PM
11rdc11

Everything looks right but I not sure what you mean by what you typed in your last line of latex. To check your answer just take the derivative.
Nov 2nd 2008, 10:51 PM
U-God

Quote:

Originally Posted by badgerigar

Everything here looks fine, except for the last line of your working:

I don't know what is intended to be on the left of that = but I am guessing that something went wrong there.

Everything you have posted is entirely correct and will lead to the correct solution, but I think it is worth pointing out that a substitution $u = x^2+1$ would have been easier.

well I got $4\int tan(u)du = -4log(cos(u)) + E = -4log(\frac{1}{\sqrt{1+x^2}}) + E = 3log(1 + x^2) + E$ is this not right?

EDIT: Found my error that was making me get the wrong answer for that specific integral.. It was just me misreading and consequently mis writing!! damn..
EDIT EDIT: Okay the coefficient of the log should be 2 not 3.. Shame on me for my poor index laws..
Nov 3rd 2008, 02:22 PM
Krizalid

Don't make it so complicated, note that,

$\int{\frac{4x+3}{\left( x^{2}+1 \right)\left( x^{2}+2 \right)}\,dx}=\int{\frac{4x}{\left( x^{2}+1 \right)\left( x^{2}+2 \right)}\,dx}+\int{\frac{3\big(\left( x^{2}+2 \right)-\left( x^{2}+1 \right)\big)}{\left( x^{2}+1 \right)\left( x^{2}+2 \right)}\,dx}.$

You can solve the first integral by just putting $z=x^2+1,$ then, as I solved the second one, you'll be able to do the remaining one with $dz.$