91Ó°ÊÓ

Statement I Integral of an even function is not always an odd function. Statement II Integral of an odd function is an even function.

The integral \(\int\left(1+x-\frac{1}{x}\right) e^{x+\frac{1}{x}} d x\) is equal to (a) \((x-1) e^{x+\frac{1}{x}}+C \quad\) (b) \(x e^{x+\frac{1}{x}}+C\) (c) \((x+1) e^{x+\frac{1}{x}}+C\) (d) \(-x e^{x+\frac{1}{x}}+C\)