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October 23rd, 2018, 12:16 PM   #1
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Plz help solve this integral...

How can we solve the following integral:

$\displaystyle \int \frac{x^{3}-1}{(x^{4}+1)(x+1)} dx$

Thx.
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October 23rd, 2018, 12:31 PM   #2
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partial fractions expansion immediately suggests itself

$\large \begin{align*}
&\frac{x^3-1}{(x^4+1)(x+1)} = \frac{x^3}{x^4+1}-\frac{1}{x+1}\\
\\
\displaystyle &\int~\frac{x^3-1}{(x^4+1)(x+1)}~dx = \\
\\
&\int ~ \frac{x^3}{x^4+1}-\frac{1}{x+1}~dx =\\
\\
&u=x^4+1,~du=4x^3\\
\\
&\dfrac 1 4 \int \dfrac{du}{u} - \ln(x+1) + C = \\
\\
&\dfrac 1 4 \ln(x^4+1) - \ln(x+1)+C


\end{align*}$
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October 23rd, 2018, 12:51 PM   #3
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Quote:
Originally Posted by romsek View Post
partial fractions expansion immediately suggests itself

$\large \begin{align*}
&\frac{x^3-1}{(x^4+1)(x+1)} = \frac{x^3}{x^4+1}-\frac{1}{x+1}\\
\\
\end{align*}$
Thx. Is there any way to do partial fraction of this expression where power 4 in denominator is there..or just observation.
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October 23rd, 2018, 02:07 PM   #4
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Quote:
Originally Posted by happy21 View Post
Thx. Is there any way to do partial fraction of this expression where power 4 in denominator is there..or just observation.
I don't understand you.
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October 23rd, 2018, 06:36 PM   #5
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I mean for partial fractions, we have some basic methods. But in this question, partial fraction was done by observation or by method (like assuming A, B, etc..in the numerator and finding the constants, etc..)?
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October 23rd, 2018, 07:09 PM   #6
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Quote:
Originally Posted by happy21 View Post
I mean for partial fractions, we have some basic methods. But in this question, partial fraction was done by observation or by method (like assuming A, B, etc..in the numerator and finding the constants, etc..)?
There is a table here that probably answers your question.
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October 23rd, 2018, 09:00 PM   #7
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Thanks for that table but only concern was about degree four polynomial in denominator which cannot be further factorized ($\displaystyle x^{4}+1$ in this question), for this particular factor, I guess we have to do partial fraction by observation.

Last edited by happy21; October 23rd, 2018 at 09:06 PM.
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October 23rd, 2018, 09:57 PM   #8
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Quote:
Originally Posted by happy21 View Post
Thanks for that table but only concern was about degree four polynomial in denominator which cannot be further factorized ($\displaystyle x^{4}+1$ in this question), for this particular factor, I guess we have to do partial fraction by observation.
If I understand the problem... what you need to do is set up

$\dfrac{x^3-1}{(x^4+1)(x+1)} = \dfrac{A x^3 + B x^3 + C x + D}{x^4+1}+\dfrac{E}{x+1}$
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October 23rd, 2018, 10:13 PM   #9
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Quote:
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. . . which cannot be further factorized . . .
$x^4 + 1 = (x^2 + \sqrt2x + 1)(x^2 - \sqrt2x + 1)$
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October 24th, 2018, 05:05 AM   #10
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Quote:
Originally Posted by skipjack View Post
$x^4 + 1 = (x^2 + \sqrt2x + 1)(x^2 - \sqrt2x + 1)$
To expand on what skipjack has done, EVERY polynomial of degree $d > 2$ can be factored into lower degree polynomials over the reals. Also EVERY polynomial of degree $d > 1$ can be factored into lower degree polynomials over the complex numbers. Hence, partial fractions works universally for problems like this.
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