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Integrate the functions(e^(tan^((-1)x)))...

Integrate the functions`(e^(tan^((-1)x)))/(1+x^2)`

Text Solution

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Given,`(e^(tan^((-1)x)))/(1+x^2)`
Put `tan^(−1)x=t`
∴`1/[1+x^2]​dx=dt`
⇒`∫[e^(tan^(−1)x​]/[1+x^2]]dx=∫e^tdt`
=`e^t+C=e^(tan^(−1)x+C`
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