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mrtuba
integral of (3x^3+2x^2+7x+13)/((x-1)^2(x^2+4)^2)
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\[\int\limits_{}^{}(3x^3+2x^2+7x+13)/((x-1)^2(x^2+4)^2)\]
Did you try partial fractions Are is partial fractions what you need help on?
i did partial fractions but its so incredibly terrible that the expanded for after cross multiplication took up to line of college rule paper written in small print. I feel like there is some simplifying i could do before doing partial fractions
It does look like it can nasty but I don't see any other way to go about it but... just because i don't see any other way doesn't mean it doesn't exist
\[\frac{3x^3+2x^2+7x+13}{(x-1)^2(x^2+4)^2}\] \[=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{Cx+D}{x^2+4}+\frac{Ex+F}{(x^2+4)^2}\] Yeah this is going to be ugly
So you didn't find those constant values for A,B,C,D,E and F right?
no i just saw the set up after cross multiplying and expanding looked so crazy that i thought i had to have missed something that could make it easier.
\[=\frac{A(x-1)(x^2+4)^2}{(x-1)^2(x^2+4)^2}+\frac{B(x^2+4)^2}{(x-1)^2(x^2+4)^2}+\frac{(Cx+D)(x-1)^2(x^2+4)}{(x-1)^2(x^2+4)^2}+\frac{(Ex+F)(x-1)^2}{(x^2+4)^2(x-1)^2}\] \[3x^3+2x^2+7x+13=A(x-1)(x^2+4)^2+B(x^2+4)^2+ \] \[(Cx+D)(x-1)^2(x^2+4)+(Ex+F)(x-1)^2 \] And then continue from here ....
yeah i went past that, just looks like this one will be ugly i guess ill just keep chugging thanks.
Sorry... I'm not able to offer an easier way :(
I don't think there is one...
Plug in X=1 to get a value, x=0 might also help
can not figure out the system of equations this leads me to
1. Carefully check your + and - signs in the original problem. You've currently an unfactorable polynomial in the numerator with a larger, higher degree polynomial in the denominator. I'd agree with @myininaya's approach. One missed sign could make it factorisable such that something could simplify/cancel out and the whole thing become much more feasible. 2. There isn't an easier way, that I can see either, you're just going to have to solve that system of equations, namely you're going to have to use both methods of combining systems of equations and making substitutions, AND using known vertical asymptote values for the denominator (in this case +1 and 2i). Yes, I'm serious, use an imaginary root here, you kind of have to. Post your system of equation here, and try those roots, see what happens (things cancel out). 3. Using that I got: A = 0 B = 1 C = 0 D = -1 E = 1 F = 1 You'll also need to make use of this fact: \[\frac{d}{dx}(\frac{1}{a} \tan^{-1}(\frac{x}{a})) = \frac{1}{x^2+a^2}\] This problem as written is do-able, but it's a lot of work and will also most likely also force you to remember some anti-derivatives which have trig function forms and make use of half-angle and double-angle formulas too. So you got a lot of work to do, better get to it eh? ;-D
Here is a summary and a conclusion to what have been above and what have not been \[ \frac{3 x^3+2 x^2+7 x+13}{(x-1)^2 \left(x^2+4\right)^2}=\frac {x+1}{\left(x^2+4\right)^2} -\frac{1}{x^2+4}+\frac{1}{( x-1)^2}\\ \int \left(\frac{x+1}{\left(x^2+4\right)^2}-\frac{1}{x^2+4}+\frac{1}{(x-1)^2}\right) \, dx=\\\frac{x-4}{8 \left(x^2+4\right)}-\frac{1}{x-1}-\frac{7 }{16} \tan ^{-1}\left(\frac{x}{2}\right)+ C \]
@eliassaab is 100% correct :-)