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

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SUMMARY

The discussion focuses on the integration of the function [(x^4+x^2+1)/2(1+x^2)]dx. A critical error identified in the integration process was the incorrect application of the arctangent function, specifically stating that ∫1/(1+x^(-2))dx does not equal tan^(-1)(1/x). The correct approach involves splitting the original expression into two parts: (x^4 + x^2)/(2(1 + x^2)) and 1/(2(1 + x^2)). This method simplifies the integration process and leads to the correct solution.

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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.
 

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