May 12th, 2017, 11:23 PM  #1 
Senior Member Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36  Integral.
How do we integrate $\displaystyle \int \frac{x^41}{x^2\sqrt{x^4+x^2+1}}dx$ ? My attempt was changing it to $\displaystyle \int \frac{x^2}{\sqrt{x^4+x^2+1}}dx\int \frac{1}{x^2\sqrt{x^4+x^2+1}}dx$, but it seems not working. 
May 13th, 2017, 10:30 AM  #2 
Senior Member Joined: May 2008 Posts: 254 Thanks: 47  
May 13th, 2017, 07:35 PM  #3 
Senior Member Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36 
I know the answer. But how to solve it?

May 14th, 2017, 02:15 PM  #4 
Senior Member Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36 
Anyone?

May 15th, 2017, 11:44 PM  #5 
Senior Member Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36 
Anyone?

May 16th, 2017, 05:07 PM  #6 
Member Joined: Apr 2015 From: USA Posts: 46 Thanks: 32 
Trig substitution works. First, complete the square: $\displaystyle\int\frac{x^41}{x^2\sqrt{x^4+2x^2+1x^2}}dx=\int\frac{x^41}{x^2\sqrt{(x^2+1)^2x^2}}dx$ Factor out and simplify: $\displaystyle\int\frac{x^41}{x^2\sqrt{(x^2+1)^2}\sqrt{1\frac{x^2}{(x^2+1)^2}}}dx=\int\frac{x^41}{x^2(x^2+1)\sqrt{1\frac{x^2}{(x^2+1)^2}}}dx=\int\frac{(x^2+1)(x^21)}{x^2(x^2+1)\sqrt{1\frac{x^2}{(x^2+1)^2}}}dx=\int\frac{\color{red}{x^ 21}}{x^2\sqrt{1\frac{x^2}{(x^2+1)^2}}}\color{red}{dx}$ Let $\displaystyle \color{blue}{\sin\theta}=\frac{\color{blue}{x}}{ \color{blue}{x^2+1}}$. Then $\displaystyle\cos\theta\,d\theta=\frac{x^2+12x^2}{(x^2+1)^2}dx=\frac{x^2+1}{(x^2+1)^2}dx=\frac{\color{red}{x^21}}{(x^2+1)^2}\color{red}{dx}$ Which means that $\displaystyle (x^2+1)^2\cos\theta\,d\theta=(\color{red}{x^21})\color{red}{dx}$ Substitute in the trig: $\displaystyle\int\frac{(x^21)}{x^2\sqrt{1\frac{x^2}{(x^2+1)^2}}}dx\Longrightarrow\int\frac{(x^2+1)^2\cos\theta}{x^2\sqrt{1\sin^2\theta}}d\theta=\int\frac{(x^2+1)^2\cos\theta}{x^2\left\cos\theta \right}d\theta$ Assuming that $\cos\theta\gt 0$, the cosines cancel: $\displaystyle\int\frac{(\color{blue}{x^2+1})^2}{\color{blue}{x}^2}d \theta$, which by another trig substitution is $\displaystyle\int\frac{1}{\color{blue}{\sin^2\theta}}d\theta$ Then $\displaystyle\int \frac{1}{\sin^2\theta}d\theta=\int \csc^2\theta\,d\theta=\cot\theta+C$ Finally, undo the trig substitution to get the final answer in $x$. 

Tags 
integral 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Double integral, repeated integral and the FTC  Jhenrique  Calculus  5  June 30th, 2015 03:45 PM 
a new discovery in integral calculus??for integral pros only  gen_shao  Calculus  2  July 31st, 2013 09:54 PM 
integral of double integral in a region E  maximus101  Calculus  0  March 4th, 2011 01:31 AM 
Prove Lower Integral <= 0 <= Upper Integral  xsw001  Real Analysis  1  October 29th, 2010 07:27 PM 
integral of double integral in a region E  maximus101  Algebra  0  December 31st, 1969 04:00 PM 