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Mathematics 20 Online
OpenStudy (anonymous):

what's the integration of 1/(x^2+1)

OpenStudy (anonymous):

You want the anti derivative?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

Trig sub \(x=\tan\theta\)

OpenStudy (anonymous):

\(dx=\sec^2\theta d\theta\)

OpenStudy (anonymous):

\[ \frac{dx}{x^2+1}=\frac{\sec^2\theta d\theta}{\sec^2\theta}=d\theta \]

OpenStudy (anonymous):

Remember that \(d\theta = \theta'dx\)

OpenStudy (anonymous):

This means \[ \theta' = \frac{1}{x^2+1} \]So \(\theta\) is the anti derivative.

OpenStudy (anonymous):

U don't have to use trig sub, 1/x^2+1 us the derivative of arctanx so it's integration is arctanx

OpenStudy (anonymous):

We just need \(\theta(x)\)

OpenStudy (anonymous):

Is*

OpenStudy (anonymous):

Since \(x=\tan(\theta)\), then \(\theta(x) = \arctan(x)\)

OpenStudy (shamil98):

wio just made it complicated xD

OpenStudy (anonymous):

@shamil98 that what iam trying to say :)

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