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find the anti derivative
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\[f'(t) = \frac{ 4 }{ 1+t^2}\]
where f(1) = 0
Hmm, do you want to do it through integration?
yes
It looks like arctan. Anyway, initially I would probably just use a trig function due to the \(1+x^2\)
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I mean \(1+t^2\).
okay so 4(arctan) + c for starting
Yeah, if you're allowed to use that fact, then it would make things easy
okay so now how do I find c?
0 = 4(arctan(1))?
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You use \(f(1)=0\).
So letting \(t=1\) and the whole thing \(=0\), you have an equation to find \(C\).
Alternatively you may use: \[ f(t) -f(1)= \int^t_1f'(x)~dx \]
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