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 October 23rd, 2018, 11:16 AM #1 Senior Member   Joined: Jan 2012 Posts: 140 Thanks: 2 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. October 23rd, 2018, 11:31 AM #2 Senior Member   Joined: Sep 2015 From: USA Posts: 2,586 Thanks: 1430 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*} Thanks from greg1313, happy21 and topsquark October 23rd, 2018, 11:51 AM   #3
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 Originally Posted by romsek 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. October 23rd, 2018, 01:07 PM   #4
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 Originally Posted by happy21 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. October 23rd, 2018, 05:36 PM #5 Senior Member   Joined: Jan 2012 Posts: 140 Thanks: 2 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..)? October 23rd, 2018, 06:09 PM   #6
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 Originally Posted by happy21 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. October 23rd, 2018, 08:00 PM #7 Senior Member   Joined: Jan 2012 Posts: 140 Thanks: 2 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 08:06 PM. October 23rd, 2018, 08:57 PM   #8
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 Originally Posted by happy21 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

$\dfrac{x^3-1}{(x^4+1)(x+1)} = \dfrac{A x^3 + B x^3 + C x + D}{x^4+1}+\dfrac{E}{x+1}$ October 23rd, 2018, 09:13 PM   #9
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 Originally Posted by happy21 . . . which cannot be further factorized . . .
$x^4 + 1 = (x^2 + \sqrt2x + 1)(x^2 - \sqrt2x + 1)$ October 24th, 2018, 04:05 AM   #10
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 Originally Posted by skipjack $x^4 + 1 = (x^2 + \sqrt2x + 1)(x^2 - \sqrt2x + 1)$
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. Tags integral, plz, solve Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post ghafarimahsa Calculus 1 September 11th, 2013 12:07 PM kaspersky0 Complex Analysis 3 October 15th, 2012 09:19 AM Ad van der ven Calculus 3 December 17th, 2011 09:53 AM malaguena Calculus 0 February 14th, 2011 01:18 AM art2 Calculus 1 September 24th, 2009 04:46 AM

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