∫∞0dx1+x3
So far I have found the indefinite integral, which is:
−16ln|x2−x+1|+1√3arctan(2x−1√3)+13ln|x+1|
Now what do I need to do in order to calculate the improper integral?
Answer
Next, simplify
F(x)=−16ln|x2−x+1|+1√3arctan2x−1√3+13ln|x+1|
=1√3arctan(2x−1√3)+13ln|x+1|−13ln√|x2−x+1|
=1√3arctan(2x−1√3)+13ln(|x+1|√|x2−x+1|).
Then
∫∞0dx1+x3=limX→∞F(X)−F(0).
Compute the limit, and you are done.
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