anonymous
  • anonymous
By expressing the following in partial fractions evaluate the given intergral.
Mathematics
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
|dw:1437741496765:dw|
UsukiDoll
  • UsukiDoll
so 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
  • UsukiDoll
now there are two forms of partial fractions: factored and non-factored 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 \[\frac{1}{x^2-1}\] we can factor \[x^2-1 = (x+1)(x-1) \] \[\frac{1}{(x+1)(x-1)}\] so for each partial fraction we have \[\frac{A}{x+1} +\frac{B}{x-1}\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

UsukiDoll
  • UsukiDoll
now we have to take the LCD which is (x-1)(x+1) \[\frac{x-1}{x-1} \cdot \frac{A}{x+1} +\frac{B}{x-1} \cdot ]\frac{x+1}{x+1}\]
anonymous
  • anonymous
I dont understand how u got from 1/(x^3+x) to 1/(x^2-1) the rest u wrote below, Im familiar with.
UsukiDoll
  • UsukiDoll
I am providing an example first anyway then we distribute the A and B \[\frac{Ax-A+Bx+B}{(x-1)(x+1)}\]
UsukiDoll
  • UsukiDoll
so we have these equations -A+B = 1 A+B = 0 we got the one from the numerator of the original equation
anonymous
  • anonymous
ah ok that makes sense.
UsukiDoll
  • UsukiDoll
so 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
  • UsukiDoll
yes I like to give a guideline first before we go to the actual problem
anonymous
  • anonymous
that example makes actual sense, my book skipped alot of steps u took.
UsukiDoll
  • UsukiDoll
oh 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 \[\large \frac{\frac{-1}{2}}{x+1} +\frac{\frac{1}{2}}{x-1}\]
UsukiDoll
  • UsukiDoll
then we take the antiderivative which is just logs
UsukiDoll
  • UsukiDoll
\[\frac{1}{2} \ln(x+1)-\frac{1}{2}\ln(x-1) +C \]
anonymous
  • anonymous
x+1 is under -1/2 dont u mean vice versa with the pos/negative signs on the 1/2s?
UsukiDoll
  • UsukiDoll
oh I missed the sign sorry
UsukiDoll
  • UsukiDoll
\[\frac{-1}{2} \ln(x+1)-\frac{1}{2}\ln(x-1) +C\]
UsukiDoll
  • UsukiDoll
so let's apply what we've done to \[\large \frac{1}{x^3+x}\]
UsukiDoll
  • UsukiDoll
can we factor the denominator?
anonymous
  • anonymous
x(x^2+1)
UsukiDoll
  • UsukiDoll
woo hoo! that means we can use single letters ^_^ should it be non-factorable we have to use Ax+B :(
UsukiDoll
  • UsukiDoll
if I make typos, please stop me. It's almost 3 am and I am typing fast
UsukiDoll
  • UsukiDoll
so anyway
anonymous
  • anonymous
Haha, okey :)
UsukiDoll
  • UsukiDoll
\[\large \frac{1}{x(x^2+1)}\]
UsukiDoll
  • UsukiDoll
so since it's factorable we use single letters \[\frac{A}{x}+\frac{B}{x^2+1}\]
UsukiDoll
  • UsukiDoll
so now we use LCD
UsukiDoll
  • UsukiDoll
\[\frac{x^2+1}{x^2+1} \cdot \frac{A}{x}+\frac{B}{x^2+1} \cdot \frac{x}{x}\]
anonymous
  • anonymous
cant I just skip that and write A(x^2+1)+bx
UsukiDoll
  • UsukiDoll
then 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
  • anonymous
okey, I will follow through then:)
UsukiDoll
  • UsukiDoll
hmmm ... the thing is we have these equations A = 1 Bx = 0 Ax^2 = 0
IrishBoy123
  • IrishBoy123
you have to use: \(\large \frac{A}{x}+\frac{Bx + C}{x^2+1}\)
IrishBoy123
  • IrishBoy123
you have a quadratic on the bottom of the second term so you need a linear term on top
UsukiDoll
  • UsukiDoll
ah geez... no problem... OH that's right x^2+1 doesn't factor thanks for the head up
UsukiDoll
  • UsukiDoll
yeah 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
  • UsukiDoll
\[\large \frac{x^2+1}{x^2+1} \cdot \frac{A}{x}+\frac{Bx + C}{x^2+1} \cdot \frac{x}{x}\]
UsukiDoll
  • UsukiDoll
there we go now we can distribute \[\large \frac{Ax^2+A+Bx^2+Cx}{x(x^2+1)}\]
UsukiDoll
  • UsukiDoll
now ...... 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
  • UsukiDoll
you'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
  • UsukiDoll
It'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
anonymous
  • anonymous
ohh okey
UsukiDoll
  • UsukiDoll
so 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
  • UsukiDoll
so we plug in A = 1, B =-1, and C = 0 back into the corrected equation shout out to @IrishBoy123 !!!!!!!! WOOOOOOOOOOOOOOOO
UsukiDoll
  • UsukiDoll
\[\large \frac{A}{x}+\frac{Bx + C}{x^2+1}\] \[\large \frac{-1}{x}+\frac{1x + 0}{x^2+1}\] \[\large \large \frac{-1}{x}+\frac{x}{x^2+1}\]
UsukiDoll
  • UsukiDoll
now we can take antiderivatives... finally after that long journey we CAN INTEGRATE!
UsukiDoll
  • UsukiDoll
first one is just ln x \[\large - \ln x+\frac{x}{x^2+1}\]
UsukiDoll
  • UsukiDoll
the second one is u-substitution
IrishBoy123
  • IrishBoy123
A = 1, B =-1
UsukiDoll
  • UsukiDoll
yeah I got that @IrishBoy123
UsukiDoll
  • UsukiDoll
A =1 , B =-1, and C = 0
UsukiDoll
  • UsukiDoll
but thanks for the verification :)
anonymous
  • anonymous
\[\frac{ -1 }{ x^2 +1} is...?\]
UsukiDoll
  • UsukiDoll
so back to u-substitution \[\large u=x^2+1 \]
UsukiDoll
  • UsukiDoll
\[\large - \ln x+\frac{x}{x^2+1} \] u-subsitution \[\large u=x^2+1 \] \[\large du = 2x dx \] \[\large \frac{1}{2}du = x dx \] \[\large - \ln x+(\frac{1}{2})\frac{1}{u} \]
UsukiDoll
  • UsukiDoll
the antiderivative of 1/u is ln u
UsukiDoll
  • UsukiDoll
\[\large - \ln x+(\frac{1}{2}) (\ln u) +C\] then subsitute u back in which was \[\large u=x^2+1 \] \[\large - \ln x+(\frac{1}{2}) (\ln x^2+1) +C\]
UsukiDoll
  • UsukiDoll
whew
UsukiDoll
  • UsukiDoll
oops miss the l l \[\large - \ln \mid x \mid +(\frac{1}{2}) (\ln \mid x^2+1 \mid) +C\]
UsukiDoll
  • UsukiDoll
so to check we just take the derivative.. it should come out to be what we had before
anonymous
  • anonymous
U spent over 30min for this question DAAMN. Thanks bro, Your a lifesaver.
UsukiDoll
  • UsukiDoll
\[\large -\frac{1}{x} + \frac{1}{2}\frac{1}{x^2+1}(2x) + \ln \mid x^2+1 \mid(0) \] then we got \[\large -\frac{1}{x}+\frac{x}{x^2+1}\]
UsukiDoll
  • UsukiDoll
which is what we had earlier... thanks XD
anonymous
  • anonymous
yup, got the correct answer as the book asked for. R u a math teacher? or tuitor of any kind?
UsukiDoll
  • UsukiDoll
math major going into teaching.
UsukiDoll
  • UsukiDoll
Partial Fractions shows up again in Ordinary Differential Equations.
anonymous
  • anonymous
I will think of you when they do :D
anonymous
  • anonymous
Alright closing this, THANK YOU <3

Looking for something else?

Not the answer you are looking for? Search for more explanations.