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

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The discussion focuses on integrating the expression [(x^4+x^2+1)/2(1+x^2)]dx, with participants noting an error in the integration process. A key point raised is that the integration of 1/(1+x^(-2)) does not equal arctan(1/x), which led to confusion in the solution. A suggestion is made to split the original expression into two simpler fractions for easier integration. This approach helps clarify the integration steps and corrects the initial misunderstanding. The conversation highlights the importance of carefully checking integration methods to ensure accurate results.
Tanishq Nandan
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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.
 

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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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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