Question

The integral \(\displaystyle\int\left(1+x-\frac1x\right)e^{x+\frac1x}\,dx\) is equal to :

• A. \((x+1)e^{x+\frac1x}+c\)
• B. \(-x\,e^{x+\frac1x}+c\)
• C. \((x-1)e^{x+\frac1x}+c\)
• D. \(x\,e^{x+\frac1x}+c\)

Hint:

Try \(x\,e^{x+1/x}\): its derivative is exactly the integrand.

Answer 1

The Correct Option is D

Let \(h=x+\frac1x\), so \(h'=1-\frac1{x^2}\).
\(\frac{d}{dx}\left(x\,e^{h}\right)=e^{h}+x\left(1-\frac1{x^2}\right)e^{h}=\left(1+x-\frac1x\right)e^{h}\), which is the integrand.
So the integral is \(x\,e^{x+\frac1x}+c\).
Answer: (4) \(x\,e^{x+\frac1x}+c\)