$\int_0^{100}\frac{e^{-x}}{x+100}dx>0.005$? $\int_0^{100}\frac{e^{-x}}{x+100}dx>0.005$?
My attempt: $$\int_0^{100}\frac{e^{-x}}{x+100}dx>\int_0^{100}\frac{e^{-x}}{200}dx=\frac{1-e^{-100}}{200}$$ A little bit error. How to amend it?

Use $e^u\ge 1+u$ to conclude 
$$
\frac1{1+u}\ge e^{-u}\implies \frac1{100+x}\ge\frac1{100}e^{-x/100}.
$$
Then, again using $1-e^{-u}\ge1-\frac1{1+u}$,
$$
\int_0^{100}\frac{e^{-x}}{100+x}dx
\ge\frac1{100} \int_0^{100} e^{-\frac{101}{100}x} dx
=\frac1{100} \frac{100}{101} \left( 1-e^{-101} \right)
\ge\frac1{101}\left(1-\frac1{102}\right)=\frac1{102}
$$
which is decidedly larger than $\frac1{200}$.
