Prove that $\int_{\ln{a}}^{\ln{b}}(1+x)(1+x^2)\left(1+x+\frac{x^2}{2}\right)\mathrm e^x\,\mathrm dx\le\frac{b^4-a^4}{4} \;?$ If $1&lt;a\leq b$, how to prove that

$$\int_{\ln{a}}^{\ln{b}}(1+x)(1+x^2)\left(1+x+\frac{x^2}{2}\right)\mathrm e^x\,\mathrm dx\le\frac{b^4-a^4}{4} \;?$$
We have 
$$ e^x = 1+x+\frac{x^2}{2}+\sum_{k=3}^\infty \frac{x^k}{k!}$$
and thus $$e^x \geq 1+x\qquad \text{ and } \qquad e^x\geq 1+x+\frac{x^2}{2}.$$ 
Now if $0\leq x\leq 1$ then $x^2\leq x$ and thus $$e^x \geq 1+x \geq 1+x^2.$$ Finally, if $x\geq 1$, let $f(x)=e^x-1-x^2$. We have $f(1)=e-2&gt;0$, let us show that $f$ is increasing for $x\geq 1$. We have $f'(x)=e^x-2x$ and thus $f'(1)=e-2&gt;0$. Moreover, $f''(x)=e^x-2&gt;0$ for every $x\geq 1$. Thus $f'$ is increasing for $x\geq 1$. It follows that $f'(x)\geq f(1)&gt;0$ for every $x\geq 1$. Thus $f$ is increasing for every $x\geq 1$. It follows that for $x\geq 1$ we have
$$0 &lt; f(1)\leq f(x) = e^x-1-x^2 \iff 1+x^2\leq e^x.$$
Thus $1+x^2\leq e^x$ for every $x \geq 0$. We finally get
$$\int_{\ln{a}}^{\ln{b}}(1+x)(1+x^2)\left(1+x+\frac{x^2}{2}\right) e^xdx \leq\int_{\ln{a}}^{\ln{b}}e^{3x}e^x dx=\frac{b^4-a^4}{4}$$
Here  x > ln(a) > 0  so each of the first three terms in the integrand is less than e^x.
So the integrand is less than e^(4x) and this can be quickly integrated.