The integral $\int\left(1+x-\frac{1}{x}\right) e^{x+\frac{1}{x}} d x$ is equal to :
$(x-1) e^{x+\frac{1}{x}}+c$
$x e^{x+\frac{1}{x}}+c$
$(x+1) e^{x+\frac{1}{x}}+c$
$-x e^{x+\frac{1}{x}}+c$
Correct Option:
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