By expressing the following in partial fractions evaluate the given intergral.

- anonymous

By expressing the following in partial fractions evaluate the given intergral.

- schrodinger

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- anonymous

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- 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

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
\[\frac{A}{x+1} +\frac{B}{x-1}\]

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## More answers

- 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

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

I am providing an example first
anyway
then we distribute the A and B
\[\frac{Ax-A+Bx+B}{(x-1)(x+1)}\]

- UsukiDoll

so we have these equations
-A+B = 1
A+B = 0
we got the one from the numerator of the original equation

- anonymous

ah ok that makes sense.

- 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

yes I like to give a guideline first before we go to the actual problem

- anonymous

that example makes actual sense, my book skipped alot of steps u took.

- 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

then we take the antiderivative which is just logs

- UsukiDoll

\[\frac{1}{2} \ln(x+1)-\frac{1}{2}\ln(x-1) +C \]

- anonymous

x+1 is under -1/2 dont u mean vice versa with the pos/negative signs on the 1/2s?

- UsukiDoll

oh I missed the sign sorry

- UsukiDoll

\[\frac{-1}{2} \ln(x+1)-\frac{1}{2}\ln(x-1) +C\]

- UsukiDoll

so let's apply what we've done to
\[\large \frac{1}{x^3+x}\]

- UsukiDoll

can we factor the denominator?

- anonymous

x(x^2+1)

- UsukiDoll

woo hoo! that means we can use single letters ^_^
should it be non-factorable we have to use Ax+B :(

- UsukiDoll

if I make typos, please stop me. It's almost 3 am and I am typing fast

- UsukiDoll

so anyway

- anonymous

Haha, okey :)

- UsukiDoll

\[\large \frac{1}{x(x^2+1)}\]

- UsukiDoll

so since it's factorable we use single letters
\[\frac{A}{x}+\frac{B}{x^2+1}\]

- UsukiDoll

so now we use LCD

- UsukiDoll

\[\frac{x^2+1}{x^2+1} \cdot \frac{A}{x}+\frac{B}{x^2+1} \cdot \frac{x}{x}\]

- anonymous

cant I just skip that and write
A(x^2+1)+bx

- 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

okey, I will follow through then:)

- UsukiDoll

hmmm ... the thing is we have these equations
A = 1
Bx = 0
Ax^2 = 0

- IrishBoy123

you have to use:
\(\large \frac{A}{x}+\frac{Bx + C}{x^2+1}\)

- IrishBoy123

you have a quadratic on the bottom of the second term so you need a linear term on top

- UsukiDoll

ah geez... no problem... OH that's right x^2+1 doesn't factor thanks for the head up

- 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

\[\large \frac{x^2+1}{x^2+1} \cdot \frac{A}{x}+\frac{Bx + C}{x^2+1} \cdot \frac{x}{x}\]

- UsukiDoll

there we go now we can distribute
\[\large \frac{Ax^2+A+Bx^2+Cx}{x(x^2+1)}\]

- 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

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

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

ohh okey

- 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

so we plug in A = 1, B =-1, and C = 0 back into the corrected equation shout out to @IrishBoy123 !!!!!!!! WOOOOOOOOOOOOOOOO

- 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

now we can take antiderivatives... finally after that long journey we CAN INTEGRATE!

- UsukiDoll

first one is just ln x
\[\large - \ln x+\frac{x}{x^2+1}\]

- UsukiDoll

the second one is u-substitution

- IrishBoy123

A = 1, B =-1

- UsukiDoll

yeah I got that @IrishBoy123

- UsukiDoll

A =1 , B =-1, and C = 0

- UsukiDoll

but thanks for the verification :)

- anonymous

\[\frac{ -1 }{ x^2 +1} is...?\]

- UsukiDoll

so back to u-substitution
\[\large u=x^2+1 \]

- 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

the antiderivative of 1/u is ln u

- 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

whew

- UsukiDoll

oops miss the l l
\[\large - \ln \mid x \mid +(\frac{1}{2}) (\ln \mid x^2+1 \mid) +C\]

- UsukiDoll

so to check we just take the derivative.. it should come out to be what we had before

- anonymous

U spent over 30min for this question DAAMN. Thanks bro, Your a lifesaver.

- 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

which is what we had earlier...
thanks XD

- anonymous

yup, got the correct answer as the book asked for.
R u a math teacher?
or tuitor of any kind?

- UsukiDoll

math major going into teaching.

- UsukiDoll

Partial Fractions shows up again in Ordinary Differential Equations.

- anonymous

I will think of you when they do :D

- anonymous

Alright closing this, THANK YOU <3

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