Integrate [(x^4+x^2+1)/2(1+x^2)]dx: Soln & Explanation

  • Thread starter Thread starter Tanishq Nandan
  • Start date Start date
  • Tags Tags
    Integration
Tanishq Nandan
Messages
122
Reaction score
5

Homework Statement


Integrate: [(x^4+x^2+1)/2(1+x^2)]dx

Homework Equations


[/B]
Integral of x^n=x^(n+1)/(n+1)
Integral of 1/(1+x^2)=arctanx

The Attempt at a Solution


I have attached my solution.All the steps seem to be correct,but the answer isn't matching,don't know why.
 

Attachments

  • 20170615_131630.jpg
    20170615_131630.jpg
    19.2 KB · Views: 371
Physics news on Phys.org
Your integration from lines 3 to 4 is not correct:
<br /> \int \frac{1}{1 + x^{-2}} \mathrm{d} x \neq \tan^{-1} \left(\frac{1}{x}\right).<br />
My suggestion is to instead split the original expression as
<br /> \frac{x^4 + x^2 + 1}{2(1 + x^{2})} = \frac{x^4 + x^2}{2(1 + x^{2})} + \frac{1}{2(1 + x^{2})}.<br />
 
Fightfish said:
My suggestion is to instead split the original expression as
x4+x2+12(1+x2)=x4+x22(1+x2)+12(1+x2).x4+x2+12(1+x2)=x4+x22(1+x2)+12(1+x2).​
Oops,why didn't I think of that?Thanks.
Fightfish said:
∫11+x−2dx≠tan−1(1x).∫11+x−2dx≠tan−1⁡(1x).​
\int \frac{1}{1 + x^{-2}} \mathrm{d} x \neq \tan^{-1} \left(\frac{1}{x}\right).
Yep,sorry.
 
Thread 'Use greedy vertex coloring algorithm to prove the upper bound of χ'
Hi! I am struggling with the exercise I mentioned under "Homework statement". The exercise is about a specific "greedy vertex coloring algorithm". One definition (which matches what my book uses) can be found here: https://people.cs.uchicago.edu/~laci/HANDOUTS/greedycoloring.pdf Here is also a screenshot of the relevant parts of the linked PDF, i.e. the def. of the algorithm: Sadly I don't have much to show as far as a solution attempt goes, as I am stuck on how to proceed. I thought...
Back
Top