For x > 0 let F(x) = ∫x^(x²) ln t dt.. Both limits of this integral move with x…

The derivative. By the Leibniz rule, $F'(x) = \ln(x^2) \cdot 2x - \ln x = 4x\ln x - \ln x = (4x - 1)\ln x$.

(a) $F'(2) = 7\ln 2 \approx 4.85$.

(b) $F'(x) = 0$ at $x = \dfrac14$ and at $x = 1$. For $0 < x < \dfrac14$ both factors are negative, so $F' > 0$. For $\dfrac14 < x < 1$, $4x - 1 > 0$ and $\ln x < 0$, so $F' < 0$. For $x > 1$ both are positive. So $F$ has a local maximum at $x = 0.25$ and a local minimum at $x = 1$.

(c) With $t\ln t - t$ as an antiderivative, $F(x) = - x^{2} + x + \left(2 x^{2} - x\right) \ln{\left(x \right)}$. At $x = \dfrac14$: $F = \frac{\ln{\left(2 \right)}}{4} + \frac{3}{16} \approx 0.3608$, that is 0.36. The value is positive even though $\ln t < 0$ on the strip: the limits run from $\dfrac14$ down to $\dfrac1{16}$, backwards, and that reverses the sign.

(d) As $x \to 0^+$, $F(x) \to 0$ (each term of the closed form tends to $0$). So $F$ rises from near $0$ to about $0.36$ on $\left(0, \dfrac14\right]$, falls to $F(1) = 0$ on $\left[\dfrac14, 1\right]$, and then rises without bound. The level $0.2$ is crossed once on each of these three stretches: 3 solutions.

(e) $F(1) = 0$ and $F'(1) = 0$. Also $F''(x) = 4\ln x + \dfrac{4x - 1}{x}$, so $F''(1) = 3$. By L'Hospital's rule twice (or by Taylor's formula), the limit is $\dfrac{F''(1)}{2} = 1.50$.