Prove $1<\frac1{e^2(e-1)}\int_e^{e^2}\frac{x}{\ln x}dxProve the following inequalities:
a) $1.43 < \int_0^1e^{x^2}dx < \frac{1+e}2$
b) $2e <\int_0^1 e^{x^2}dx+\int_0^1e^{2-x^2}dx<1+e^2$
c) $1<\frac1{e^2(e-1)}\int_e^{e^2}\frac{x}{\ln x}dx<e/2$
Source: http://www.sms.edu.pk/downloads/prceedingsfordae1/1_intinequal.pdf

We know $\sqrt{\pi}/2=\int_0^\infty e^{x^2}dx>\int_0^1 e^{x^2}dx$, and according to Google calculator (https://www.google.com/search?q=sqrt(pi)%2F2&oq=sqrt(pi)%2F2&aqs=chrome..69i57j0.3176j0j7&sourceid=chrome&espv=210&es_sm=93&ie=UTF-8 ) $\sqrt{\pi}/2\approx 0.88622692545$, while $\frac{1+e}2>1$. But $1.43$ doesn't look right. Maybe there's a typo in the pdf file.

for c use mean value theorem for integral calculus.Ie. take $f(x)=\frac{x}{ln x}$ in the interval $[e,e^2]$.
since maximum value of the function in the given interval is $e$ and $\frac{e^2}{2}$ at $x=e$    and  $x=e^2$  using maxima minima test.Hence applying meanvalue theorem we get
$$ e<\frac{\int_e^{e^2} \frac{x}{ln x}}{e^2-e}<\frac{e^2}{2}$$
rearranging we get $$1<\frac{\int_e^{e^2} \frac{x}{ln x}}{e^2(e-1)}<\frac{e}{2}$$
