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anonymous
 one year ago
By expressing the following in partial fractions evaluate the given intergral.
anonymous
 one year ago
By expressing the following in partial fractions evaluate the given intergral.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437741496765:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2so first we have to take partial fractions before we can take the antiderivative. we can ignore the integral symbol for the first half of this problem

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2now there are two forms of partial fractions: factored and nonfactored we have to look closely in the denominator.. if we can factor something out. our numerator will just be a single letter like A for example suppose we are given this fraction \[\frac{1}{x^21}\] we can factor \[x^21 = (x+1)(x1) \] \[\frac{1}{(x+1)(x1)}\] so for each partial fraction we have \[\frac{A}{x+1} +\frac{B}{x1}\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2now we have to take the LCD which is (x1)(x+1) \[\frac{x1}{x1} \cdot \frac{A}{x+1} +\frac{B}{x1} \cdot ]\frac{x+1}{x+1}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I dont understand how u got from 1/(x^3+x) to 1/(x^21) the rest u wrote below, Im familiar with.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2I am providing an example first anyway then we distribute the A and B \[\frac{AxA+Bx+B}{(x1)(x+1)}\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2so we have these equations A+B = 1 A+B = 0 we got the one from the numerator of the original equation

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ah ok that makes sense.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2so now we solve these system of equations through elimination method. That would give us 2B=1 or B =1/2 so we plug that value in one of the equations \[A+\frac{1}{2}=0\] so our \[A = \frac{1}{2}\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2yes I like to give a guideline first before we go to the actual problem

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that example makes actual sense, my book skipped alot of steps u took.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2oh dear T_T glad I came lol so we plug in A = 1/2 and B = 1/2 back into the equation before we took the LCD \[\large \frac{\frac{1}{2}}{x+1} +\frac{\frac{1}{2}}{x1}\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2then we take the antiderivative which is just logs

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2\[\frac{1}{2} \ln(x+1)\frac{1}{2}\ln(x1) +C \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0x+1 is under 1/2 dont u mean vice versa with the pos/negative signs on the 1/2s?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2oh I missed the sign sorry

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2\[\frac{1}{2} \ln(x+1)\frac{1}{2}\ln(x1) +C\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2so let's apply what we've done to \[\large \frac{1}{x^3+x}\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2can we factor the denominator?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2woo hoo! that means we can use single letters ^_^ should it be nonfactorable we have to use Ax+B :(

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2if I make typos, please stop me. It's almost 3 am and I am typing fast

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2\[\large \frac{1}{x(x^2+1)}\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2so since it's factorable we use single letters \[\frac{A}{x}+\frac{B}{x^2+1}\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2\[\frac{x^2+1}{x^2+1} \cdot \frac{A}{x}+\frac{B}{x^2+1} \cdot \frac{x}{x}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0cant I just skip that and write A(x^2+1)+bx

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2then we distribute \[\frac{Ax^2+A+Bx}{x(x^2+1)}\] I don't know if it's a good idea though. skipping steps at this topic is not recommended in my opinion

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okey, I will follow through then:)

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2hmmm ... the thing is we have these equations A = 1 Bx = 0 Ax^2 = 0

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0you have to use: \(\large \frac{A}{x}+\frac{Bx + C}{x^2+1}\)

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0you have a quadratic on the bottom of the second term so you need a linear term on top

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2ah geez... no problem... OH that's right x^2+1 doesn't factor thanks for the head up

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2yeah x^2+1 doesn't factor so Bx+C so \[x^3+x \rightarrow (x)(x^2+1) \] a portion is factored. we can take an x out but we can't do anything to x^2+1

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2\[\large \frac{x^2+1}{x^2+1} \cdot \frac{A}{x}+\frac{Bx + C}{x^2+1} \cdot \frac{x}{x}\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2there we go now we can distribute \[\large \frac{Ax^2+A+Bx^2+Cx}{x(x^2+1)}\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2now ...... we just have one value from the original equation \[\large \frac{1}{x^3+x} \] So we have three equations A = 1 C =0 A+B = 0

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2you're probably wondering why 1? why is the rest of the equations = 0 it's because our numerator is just that number if we have something like 1 + 2x on the numerator the C = 2 due to the fact that we had Cx after we did partial fractions

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2It's like this A= 1 Cx = 0 Ax^2+Bx^2 = 0 because there is no number next to the x and no number next to the x^2 so they are set to 0... we don't have to write the variable of x's for these letter equations

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2so we know that A = 1 C = 0 A+B = 0 we know C and A so we solve for B 1+B=0 B = 1

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2so we plug in A = 1, B =1, and C = 0 back into the corrected equation shout out to @IrishBoy123 !!!!!!!! WOOOOOOOOOOOOOOOO

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2\[\large \frac{A}{x}+\frac{Bx + C}{x^2+1}\] \[\large \frac{1}{x}+\frac{1x + 0}{x^2+1}\] \[\large \large \frac{1}{x}+\frac{x}{x^2+1}\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2now we can take antiderivatives... finally after that long journey we CAN INTEGRATE!

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2first one is just ln x \[\large  \ln x+\frac{x}{x^2+1}\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2the second one is usubstitution

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2yeah I got that @IrishBoy123

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2A =1 , B =1, and C = 0

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2but thanks for the verification :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ 1 }{ x^2 +1} is...?\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2so back to usubstitution \[\large u=x^2+1 \]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2\[\large  \ln x+\frac{x}{x^2+1} \] usubsitution \[\large u=x^2+1 \] \[\large du = 2x dx \] \[\large \frac{1}{2}du = x dx \] \[\large  \ln x+(\frac{1}{2})\frac{1}{u} \]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2the antiderivative of 1/u is ln u

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2\[\large  \ln x+(\frac{1}{2}) (\ln u) +C\] then subsitute u back in which was \[\large u=x^2+1 \] \[\large  \ln x+(\frac{1}{2}) (\ln x^2+1) +C\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2oops miss the l l \[\large  \ln \mid x \mid +(\frac{1}{2}) (\ln \mid x^2+1 \mid) +C\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2so to check we just take the derivative.. it should come out to be what we had before

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0U spent over 30min for this question DAAMN. Thanks bro, Your a lifesaver.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2\[\large \frac{1}{x} + \frac{1}{2}\frac{1}{x^2+1}(2x) + \ln \mid x^2+1 \mid(0) \] then we got \[\large \frac{1}{x}+\frac{x}{x^2+1}\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2which is what we had earlier... thanks XD

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yup, got the correct answer as the book asked for. R u a math teacher? or tuitor of any kind?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2math major going into teaching.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2Partial Fractions shows up again in Ordinary Differential Equations.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I will think of you when they do :D

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alright closing this, THANK YOU <3
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