Question

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 1

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

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