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Integration by Partial Fractions
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[QUOTE="FaraDazed, post: 4752163, member: 438851"] [h2]Homework Statement [/h2] Find the indefinite integral of the below, using partial fractions. [tex] \frac{4x^2+6x-1}{(x+3)(2x^2-1)} [/tex] [h2]Homework Equations[/h2] ?[h2]The Attempt at a Solution[/h2] First I want to say there is probably a much easier and quicker way to get around certain things I have done but I have just done it the only way I could see (I always seem to go the long way around). So if I am correct then I would really appreciate some tips on how to do it quicker (for in exams) and also if I am incorrect to point out my mistakes. Thanks :) first I set up the partial fractions. [itex] \frac{4x^2+6x-1}{(x+3)(2x^2-1)}=\frac{A}{x+3}+\frac{Bx+C}{2x^2-1} \\ \rightarrow \,\,\, 4x^2+6x-1=A(2x^2-1)+(Bx+C)(x+3) \\ 4x^2+6x-1=2Ax^2-A+Bx^2+3Bx+Cx+3C [/itex] And then equated the coefficients: For X^2: [itex]4=2A+B[/itex] For X^1: [itex]6=3B+C[/itex] For X^0: [itex]-1=3C-A[/itex] Then what I did what multiply the third equation above by 2 to get [itex]-2=6C-2A[/itex] and then added it to the first equation to get [itex]2=6C+B[/itex] and solved for B and substituted it into the second equation which resulted in: [itex] 6=3(2-6C)+C\\ 6=6-11C \\ 0=-11C \\ ∴ C=0 [/itex] And then knowing C=0 I found A=1 and then that B=2 . Which then leads to: [itex] \frac{4x^2+6x-1}{(x+3)(2x^2-1)}=\frac{1}{x+3}+\frac{2x}{2x^2-1} \\ \frac{4x^2+6x-1}{(x+3)(2x^2-1)}=\frac{1}{x+3}+\frac{2x}{2x^2-1}\\ \int \frac{4x^2+6x-1}{(x+3)(2x^2-1)} \,\, dx = \ln{|x+3|} + \frac{1}{2} \ln{|2x^2-1|} [/itex] I think I may have done something wrong. I had to use an integral calculator online to find the integral of [itex]\frac{2x}{2x^2-1}[/itex] as I had no idea how to do it. Which kind of defeats the point. [/QUOTE]
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