Prove: $\int_{0}^{1} \frac{e^x}{x^{2}+1}dx\le e -1 $ Prove: $\int_{0}^{1} \frac{e^x}{x^{2}+1}dx\le e -1 $
this isn't really a computable integral, But the only idea I have in mind is that:
derivative from $x=0$ to $x=1$ of:  $\frac{d}{dx}\left({e^x}-{x})\right) = e -1 $, but I would still have to compute $\frac{e^x}{x^{2}+1}$,
because I'm not sure the derivatives will be sufficient in order to prove the inequality.
 A: Because of $\frac{1}{x^2+1}\leq 1$ for all $x$, we have $$\int_0^1 \frac{e^x}{x^2+1}dx\leq \int_0^1 e^x dx= e-1.$$
The inequality $\frac{1}{x^2+1}\leq 1$ follows from $x^2+1\geq 1$. This inequality is not very sharp though: The integrals equals, roughly, 1.271 whereas the upper bound is $e-1=1.718$.
A: A better bound can be achieved by applying the Integral Chebyshev inequality to $f(x) = \frac{1}{1+x^2}$ and $g(x) = e^{x}$. Since $f$ and $g$ are of opposite monotonicity on $[0, 1]$,
$$
 \int_0^1 \frac{e^x}{1+x^2} \, dx \le \int_0^1 \frac{1}{1+x^2} \, dx \cdot \int_0^1 e^x \, dx = \frac{\pi}{4} (e-1) \approx 1.3495 \, .
$$
A: The most direct approach to this question is Triangle Inequality for Integrals. Below I provided a helpful link.
https://proofwiki.org/wiki/Triangle_Inequality_for_Integrals
Let us start by applying the inequality: $$\Biggl \lvert \int_{0}^{1} \frac{e^x}{x^2+1}dx\Biggl\rvert \space \leq \space \int_{0}^{1} \Biggl| \frac{e^x}{x^2+1}\Biggl|dx$$ Since we know that the function $\frac{e^x}{x^2+1}$ is positive-valued on the real line, we have: $$\int_{0}^{1} \Biggl| \frac{e^x}{x^2+1}\Biggl|dx = \int_{0}^{1} \frac{e^x}{x^2+1}dx$$ By the help of a simple algebraic fact $\frac{1}{1+x^2} \leq 1 \space \forall \space x \in \Bbb R$, it is easy to deduce that: $$\int_{0}^{1} \frac{e^x}{x^2+1}dx \space  \leq \space \int_{0}^{1} e^x dx=[e^x]^{1}_{0}=e-1 $$
In conclusion: $$\bbox[yellow] {\int_{0}^{1} \frac{e^x}{x^2+1} dx \space \leq \space e-1 }$$
