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anonymous
 5 years ago
How does the integral of (du/(u^2+4)) become 1/2 arctan (x+1/2) + C?
anonymous
 5 years ago
How does the integral of (du/(u^2+4)) become 1/2 arctan (x+1/2) + C?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Using Trig substituion

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0u=tan(x) du=sec(x)^2 du

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0How did u become over two, what happened to the four in the equation?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0substituting u=2tan(x) du=2sec(x)^2dx so u^2+4 is 4tan(x)^2+4 4(tan(x)^2 +1) \[1/4*2\int\limits \sec(x)^2/\sec(x)^2 d(x)\]
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