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Question

Mathematics Question on integral

∫dxex+e−x∫\frac{dx}{e^x+e^{-x}} is equal to

A

tan−1(ex)+Ctan^{-1}(e^x)+C

B

tan−1(e−x)+Ctan^{-1}(e^{-x})+C

C

tan−1(ex−e−x)+Ctan^{-1}(e^x-e^{-x})+C

D

log(ex+e−x)+Clog(e^x+e^{-x})+C

Answer

tan−1(ex)+Ctan^{-1}(e^x)+C

Explanation

Solution

The correct answer is A:=tan−1(ex)+C=tan^{-1}(e^x)+C
Let I=∫dxex+e−xdx=∫exe2x+1dxI=∫\frac{dx}{e^x+e^{-x}}dx=∫\frac{e^x}{e^{2x}+1}dx
Also,let ex=t⇒exdx=dte^x=t⇒e^x dx=dt
∴I=∫dt1+t2∴I=∫\frac{dt}{1+t^2}
=tan−1+C=tan^{-1}+C
=tan−1(ex)+C=tan^{-1}(e^x)+C
Hence,the correct Answer is A.