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Find \(\int \frac{e^{-cot^{-1}⁡x}}{1+x^2}\).(a) \(e^{-cot^{-1}⁡x}+C\)(b) \(e^{-2cot^{-1}⁡x}+C\)(c) \(e^{-tan^{-1}⁡x}+C\)(d) \(e^{-cot^1⁡2x}+C\)This question was addressed to me during a job interview.Enquiry is from Methods of Integration-1 in section Integrals of Mathematics – Class 12

Answer» CORRECT option is (a) \(E^{-cot^{-1}⁡x}+C\)

The explanation: Let \(-cot^{-1}⁡x\)=t

Differentiating w.r.t x, we GET

–\(\left (-\frac{1}{1+x^2}\right )dx=dt\)

\(\frac{1}{1+x^2} dx=dt\)

\(\INT \frac{e^{-cot^{-1}}x}{1+x^2} dx=\int e^t \,dt\)

=e^t

Replacing t with -cot^-1x, we get

\(\int \frac{e^{-cot^{-1}}x}{1+x^2} dx=e^{-cot^{-1}}x+C\)


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