Calculus 5 - by Vlad Vasilescu (W)

\int\limits_1^\sqrt{2} e^{m \cdot x^2 + ln(x)} dx = \int\limits_1^\sqrt{2} e^{m \cdot x^2} \cdot e^{ln(x)} dx = \int\limits_1^\sqrt{2} $x \cdot e^{m \cdot x^2} dx =\left. \frac {e^{m \cdot x^2}}{2 \cdot m} \right|_1^{\sqrt{2}} = \frac {e^m \cdot (e^m-1)}{2 \cdot m} = \frac {1}{m}$ . $\Rightarrow e^m \cdot (e^m-1) = 2$ . We make the notation $e^m = t$ and we get $t_{1} = 2$ and $t_{2} = -1$ .

• for $t = 2$ we get $m = ln(2)$
• for $t = -1$ there are no solutions for m

Hence , our answer will be $m = ln(2) \approx \boxed{0.69}$