Why does the improper integral $\int_{-1}^{1}\frac{e^x}{(x+1)}\mathrm d x$ diverge?

Question

I'm having trouble finding a function to compare it to on [-1,0]...

Answer 1

Hint: $e^x$ is a bounded function which has no roots.

Answer 2

By the David Mitra's hint!

$$e^x\ge\dfrac{1}{e}\quad\text{ if}\quad x\ge-1$$ $$\int_{-1}^1\dfrac{e^x}{1+x}\ge\int_{-1}^1\dfrac{1}{e(1+x)}=\left.\dfrac{1}{e}\ln(1+x)\right|_{-1}^1$$ Clearly this is divergent.