Thread: indefnite integral
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Old December 1st, 2008, 07:49 PM
Chop Suey Chop Suey is offline
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Well, one way is to notice that:
\int \frac{2x}{\sqrt{1+(x^2)^2}}~dx

Try a trigonometric substitution such as x^2=\tan{u} which yields 2x~dx = \sec^2{u}~du. You eventually end up with:

\int \sec{u}~du = \ln{|\sec{u}+\tan{u}|}+C

Don't forget to back-substitute.
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