Solving Integrals: \int\frac{xdx}{1+x^4}

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SUMMARY

The integral \(\int\frac{xdx}{1+x^4}\) was solved using the substitution \(u=x^2\), leading to the transformation \(\frac{1}{2}\int\frac{du}{1+u^2}\). The final result, \(\frac{1}{2} \arctan(x^2) + c\), is confirmed as correct. An alternative expression, \(\frac{1}{2} \arctan(1+x^2) + c\), is suggested for simplification. The discussion highlights effective u-substitution techniques in integral calculus.

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Echo 6 Sierra
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OK, for the bad mamma-jamma:
[tex]\int\frac{xdx}{1+x^4}[/tex] I picked [tex]u=x^2[/tex] and made du=2xdx which makes 1/2du=xdx.

Now I have: [tex]\displaystyle{\frac{1}{2}}[/tex][tex]\int\frac{du}{1+u^2}[/tex]

If the almighty green-bottled lager from Holland has been good to me, I get:

[tex]\frac{1}{2} \arctan x^2+c[/tex]

Is this: a) correct, b)good enough or c)can this be tweaked?

So help me, if I get at least a B in Calculus I will be near Spiritual Creaminess.
 
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Yes it's correct.
 


First of all, great job on solving the integral! Your approach of using u-substitution and picking u=x^2 was very effective. Your final answer of 1/2 arctan x^2 + c is indeed correct and can be considered good enough. However, if you want to tweak it a bit, you can also express it as 1/2 arctan (1+x^2) + c. This is just a different way of writing the same answer and may be considered more simplified. Overall, your work shows a strong understanding of integration techniques and I have no doubt that you will do well in Calculus. Keep up the good work!
 

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