Consider the equation $\displaystyle\int_{1}^{e} \frac{(\log_e x)^{1/2}}{x\left(a-(\log_e x)^{3/2}\right)^{2}}\,dx = 1, \quad a \in (-\infty,0)\cup(1,\infty).$

Consider the equation ∫₁ᵉ ((logₑ x)^(1/2))/(x(a-(logₑ x)^(3/2))²) dx = 1, a ∈ (-∞,0)∪(1,∞). $\int_1^e \dfrac{(\log_e x)^{1/2}}{x\left(a-(\log_e x)^{3/2}\right)^{2}}\,dx = 1$ $a \in (-\infty,0)\cup(1,\infty)$ $\text{To find: which statements about } a \text{ are correct}$ $\text{Plan: put } u=(\log_e x)^{3/2},\ \text{integrate, then solve for } a$ $u=(\log_e x)^{3/2} \Rightarrow du = \dfrac{3}{2}(\log_e x)^{1/2}\dfrac{dx}{x}$ $\dfrac{(\log_e x)^{1/2}}{x}\,dx = \dfrac{2}{3}\,du;\qquad x=1\to u=0,\ \ x=e\to u=1$ $I = \dfrac{2}{3}\int_0^1 \dfrac{du}{(a-u)^2}$ $\int \dfrac{du}{(a-u)^2} = \dfrac{1}{a-u}$ $I = \dfrac{2}{3}\left[\dfrac{1}{a-1} - \dfrac{1}{a}\right] = \dfrac{2}{3}\cdot\dfrac{1}{a(a-1)}$ $\dfrac{2}{3}\cdot\dfrac{1}{a(a-1)} = 1 \Rightarrow a(a-1)=\dfrac{2}{3}$ $3a^2 - 3a - 2 = 0$ $a = \dfrac{3\pm\sqrt{9+24}}{6} = \dfrac{3\pm\sqrt{33}}{6}$ $a_1=\dfrac{3+\sqrt{33}}{6}\approx 1.46>1\ \checkmark,\qquad a_2=\dfrac{3-\sqrt{33}}{6}\approx -0.46<0\ \checkmark$ $(1)\ \text{FALSE}\quad (2)\ \text{FALSE}\quad (3)\ \text{TRUE}\quad (4)\ \text{TRUE}$ $\boxed{(3)\ \text{and}\ (4)}$