Indefinite Integration Challenge to the MMF Community

Oct 2011
224
0
India ????
Here are some challenging problems which you may enjoy.

Problem 1

Evaluate \(\displaystyle \int \frac{1-x}{(1+x)\sqrt{x+x^2+x^3}}dx\)

Problem 2

Evaluate \(\displaystyle \int\frac{2-x^2}{(1+x)\sqrt{1-x^2}}dx\)

Problem 3

Evaluate \(\displaystyle \int \frac{x^7-24x^4-4x^2+8x-8}{x^8+6x^6+12x^4+8x^2}dx\)

I can provide hints if required.

Have Fun!
 

agentredlum

Math Team
Jul 2011
3,372
234
North America, 42nd parallel
THIS POST IS OFF TOPIC

NICE AVATAR!!!!

-agent
 

mathbalarka

Math Team
Mar 2012
3,871
86
India, West Bengal
The first one doesn't seems elementary to me, is it the case? The way I evaluated, I think it requires the elliptic integrals.

Balarka
.
 
Oct 2011
224
0
India ????
Everything is elementary. The first one is merely

\(\displaystyle -2\arctan \sqrt{x+\frac{1}{x}+1}+C\)

And I don't see anything wrong with my post.
 

greg1313

Forum Staff
Oct 2008
8,008
1,174
London, Ontario, Canada - The Forest City
2. \(\displaystyle \int\frac{2\,-\,x^2}{(1\,+\,x)\sqrt{1\,-\,x^2}}\,dx\)

\(\displaystyle \sin(\theta)\,=\,x,\,\cos(\theta)\,d\theta\,=\,dx\)

\(\displaystyle \int\frac{\[2\,-\,\sin^2(\theta)\]\cos(\theta)}{\[1\,+\,\sin(\theta)\]\cos(\theta)}\,d\theta\,=\,\int\frac{1}{1\,+\sin(\theta)}\,+\,1\,-\,\sin(\theta)\,d\theta\,=\,\int\frac{1}{1\,+\,\sin(\theta)}\,d\theta\,+\,\theta\,+\,\cos(\theta)\,+\,C\)

\(\displaystyle \int\frac{1}{1\,+\,\sin(\theta)}\,d\theta\,\Leftrightarrow\,t\,=\,\tan\(\frac{\theta}{2}\)\,\Leftrightarrow\,\int\frac{2}{(1\,+\,t)^2}\,dt\,=\,-2(1\,+\,t)^{-1}\,+\,C\,=\,-2\frac{1}{1\,+\,\tan\(\frac{\theta}{2}\)}\,+\,C\)

\(\displaystyle -2\frac{1}{1\,+\,\tan\(\frac{\theta}{2}\)}\,+\,\theta\,+\,\cos(\theta)\,+\,C\,=\,-2\,\cdot\,\frac{1\,+\,\sqrt{1\,-\,x^2}}{1\,+\,x\,+\,\sqrt{1\,-\,x^2}}\,+\,\sin^{\small{-1}}(x)\,+\,\sqrt{1\,-\,x^2}\,+\,C\,=\,\int\frac{2\,-\,x^2}{(1\,+\,x)\sqrt{1\,-\,x^2}}\,dx\)

I used the Weierstrass Substitution.
 

CRGreathouse

Forum Staff
Nov 2006
16,046
936
UTC -5
Etyucan said:
And I don't see anything wrong with my post.
If this is a response to agentredlum, I believe he was suggesting that *his* post was off-topic.
 

agentredlum

Math Team
Jul 2011
3,372
234
North America, 42nd parallel
CRGreathouse said:
Etyucan said:
And I don't see anything wrong with my post.
If this is a response to agentredlum, I believe he was suggesting that *his* post was off-topic.
Yes, MY post was off topic but i couldn't resist complimenting you about Seraph, from Matrix, as seen by NEO codevision, avatar.

Seraph protects that which matters most.

:D
 
Oct 2011
224
0
India ????
CRGreathouse said:
If this is a response to agentredlum, I believe he was suggesting that *his* post was off-topic.
OH! I am sorry. I thought he meant about my post.
 
May 2011
501
6
Hey E. For the first one it is funny, if one runs it through Maple or Mathematica it returns this long convoluted Elliptic monstrosity.

I got something that is different than your arctan, but equivalent.

\(\displaystyle -\sin^{-1}\left(\frac{x^{2}+1}{(x+1)^{2}}\right)-\frac{\pi}{2}\)

Taking the derivative of this returns the integral you posted.

I rewrote it as \(\displaystyle \frac{1-x}{x(1+x)\sqrt{\frac{x^{2}+x+1}{x}}}\)

Made the sub \(\displaystyle x=\frac{1+t}{1-t}, \;\ dx=\frac{2}{(t-1)^{2}}dt\)

\(\displaystyle -2\int \frac{t}{\sqrt{(1-t^{2})(t^{2}+3)}}dt\)

\(\displaystyle -2\int\frac{t}{\sqrt{1-\left(\frac{t^{2}+1}{2}\right)^{2}}}dt\)

This is an arcsin integral which can be seen to evaluate to \(\displaystyle -\sin^{-1}\left(\frac{t^{2}+1}{2}\right)+C\)

Resubbing gives the final result along with \(\displaystyle C=\frac{-\pi}{2}\)
 
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