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: 2,531 Thanks: 1390 
partial fractions expansion immediately suggests itself $\large \begin{align*} &\frac{x^31}{(x^4+1)(x+1)} = \frac{x^3}{x^4+1}\frac{1}{x+1}\\ \\ \displaystyle &\int~\frac{x^31}{(x^4+1)(x+1)}~dx = \\ \\ &\int ~ \frac{x^3}{x^4+1}\frac{1}{x+1}~dx =\\ \\ &u=x^4+1,~du=4x^3\\ \\ &\dfrac 1 4 \int \dfrac{du}{u}  \ln(x+1) + C = \\ \\ &\dfrac 1 4 \ln(x^4+1)  \ln(x+1)+C \end{align*}$ 
October 23rd, 2018, 11:51 AM  #3 
Senior Member Joined: Jan 2012 Posts: 140 Thanks: 2  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 
Senior Member Joined: Sep 2015 From: USA Posts: 2,531 Thanks: 1390  
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  
Senior Member Joined: Sep 2015 From: USA Posts: 2,531 Thanks: 1390  Quote:
 
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  
Senior Member Joined: Sep 2015 From: USA Posts: 2,531 Thanks: 1390  Quote:
$\dfrac{x^31}{(x^4+1)(x+1)} = \dfrac{A x^3 + B x^3 + C x + D}{x^4+1}+\dfrac{E}{x+1}$  
October 23rd, 2018, 09:13 PM  #9 
Global Moderator Joined: Dec 2006 Posts: 20,931 Thanks: 2207  
October 24th, 2018, 04:05 AM  #10 
Senior Member Joined: Sep 2016 From: USA Posts: 635 Thanks: 401 Math Focus: Dynamical systems, analytic function theory, numerics  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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