Integrating Rational Functions by Partial Fractions question

[hank]

Ok, here's the problem..

S dx/(x^3 + x)

So I split the problem up like this:

A/x + B/(x^2 + 1) = 1/(x(x^2 + 1))
= A(x^2 + 1) + Bx = 1
When x = 0: A = 1

Because there is no x left on the right side, B must = 0, right?

So 1 + B = 0, therefore B = -1.

That gives me

S dx/x = S dx/(x^2 + 1)
= ln|x| - arctan(x) + C

But according to the integrator, it should be:
= ln|x| - 1/2 ln(x^2 + 1) + C

What am I missing here?

 

[galactus]

You have your PF set up a little incorrectly. It's Bx, not B, because of the
quadratic.

\(\displaystyle \L\\\int\frac{1}{x^{3}+x}dx\)


Factor the denominator:

\(\displaystyle \L\\\frac{1}{\underbrace{x}_{\text{A}}\underbrace{(x^{2}+1)}_{\text{Bx}}}\)

You now have:

\(\displaystyle \L\\\frac{A}{x}+\frac{Bx}{x^{2}+1}=1\)

Now, can you finish?.

 

[soroban]

Re: Integrating Rational Functions by Partial Fractions ques

Hello, Hank!

\(\displaystyle \L\int\frac{dx}{x^3\,+\,x}\)

The set up is: \(\displaystyle \L\:\frac{1}{x(x^2\,+\,1)} \;=\;\frac{A}{x^{\,}}\,+\,\frac{Bx\,+\,C}{x^2\,+\,1}\)


Then: \(\displaystyle \:1 \;=\;A(x^2+1)\,+\,Bx^2\,+\,Cx\)

Let \(\displaystyle x\,=\,0:\;\;1\;=\;A(1)\,+\,B(0)\,+\,C(0)\;\;\Rightarrow\;\;\fbox{A\,=\,1}\)

Let \(\displaystyle x\,=\,1:\;\;1\;=\;A(2)\,+\,B(1)\,+\,C(1)\;\;\Rightarrow\;\;B\,+\,C\:=\:-1\;\) [1]

Let \(\displaystyle x\,=\,-1:\;\;1\;=\;A(2)\,+\,B(1)\,+\,C(-1)\;\;\Rightarrow\;\;B\,-\,C\:=\:-1\;\) [2]

Add [1] and [2]: \(\displaystyle \:2B\,=\,-2\;\;\Rightarrow\;\;\fbox{B\,=\,-1}\;\;\Rightarrow\;\;\fbox{C\,=\,0}\)


The decomposition is: \(\displaystyle \L\:\frac{1}{x(x^2\,+\,1} \;=\;\frac{1}{x^{\,}}\,-\,\frac{x}{x^2\,+\,1}\)

 

[galactus]

Oops, I forgot my C. blush

It still works out the same, but only because C=0.