Consider the equation

$$\int_1^e \frac{(\log_e x)^{1/2}}{x\left(a - (\log_e x)^{3/2}\right)^2} dx = 1, \quad a \in (-\infty, 0) \cup (1, \infty).$$

Which of the following statements is/are TRUE?

Write $$\ln x$$ instead of $$\log_e x$$ to shorten the symbols.

The equation is
$$\int_{1}^{e} \frac{(\ln x)^{1/2}}{x\left(a-(\ln x)^{3/2}\right)^2}\,dx = 1,$$ with $$a \in (-\infty,0)\cup(1,\infty).$$

Step 1 : Suitable substitution
Let$$t=(\ln x)^{3/2}\,.$$

Then $$\ln x = t^{2/3}\quad\Longrightarrow\quad x = e^{\,t^{2/3}}$$ and hence $$dx = e^{\,t^{2/3}}\cdot\frac{2}{3}\,t^{-1/3}\,dt.$$ (We used $$\frac{d}{dt}\bigl(t^{2/3}\bigr)=\frac{2}{3}t^{-1/3}$$.)

Step 2 : Transform the integrand
The numerator becomes $$(\ln x)^{1/2}=t^{1/3},$$ while the denominator equals $$x\bigl(a-(\ln x)^{3/2}\bigr)^2=e^{\,t^{2/3}}(a-t)^2.$$ Therefore $$\frac{(\ln x)^{1/2}}{x\left(a-(\ln x)^{3/2}\right)^2}dx =\frac{t^{1/3}}{e^{\,t^{2/3}}(a-t)^2}\;\Bigl[e^{\,t^{2/3}}\cdot\frac{2}{3}t^{-1/3}\,dt\Bigr] =\frac{2}{3}\,\frac{1}{(a-t)^2}\,dt.$$ The factor $$e^{\,t^{2/3}}$$ cancels out completely.

Step 3 : New limits
When $$x=1,\; \ln 1 =0\;\Longrightarrow\; t=0.$|
When $$x=e,\; $$\ln$$ e =1\;\Longrightarrow\; t=1.$$

Hence the original integral reduces to $$$$\frac{2}{3}$$$$\int_{0}^{1}\frac{dt}{(a-t$$)^2}=1.$$

Step 4 : Evaluate the simple integral
Recall $$$$\int$$$$\frac{dt}{(a-t)^2}=\frac{1}{a-t}$$+C,$$ since $$$$\frac{d}{dt}$$\Bigl($$\frac{1}{a-t}$$\Bigr)=$$\frac{1}{(a-t)^2}$$.$$
Thus $$$$\int_{0}^{1}\frac{dt}{(a-t$$)^2} =\Bigl[$$\frac{1}{a-t}$$\Bigr]_{0}^{1} =$$\frac{1}{a-1}-\frac{1}{a}$$.$$ Therefore $$$$\frac{2}{3}$$$$\left$$($$\frac{1}{a-1}-\frac{1}{a}$$$$\right$$)=1.$$

Step 5 : Solve for $$a$$
Combine the fractions: $$$$\frac{1}{a-1}-\frac{1}{a}=\frac{a-(a-1)}{a(a-1)}=\frac{1}{a(a-1)}$$.$$ Insert this into the equation: $$$$\frac{2}{3}$$$$\cdot$$$$\frac{1}{a(a-1)}$$=1 \;\Longrightarrow\; $$\frac{2}{3a(a-1)}$$=1 \;\Longrightarrow\; 2=3a(a-1).$$ Expand: $$3a^{2}-3a-2=0.$$ Solve the quadratic: $$a=$$\frac{3\pm\sqrt{9+24}$$}{6} =$$\frac{3\pm\sqrt{33}$$}{6}.$$

Step 6 : Check domain
Compute the two roots approximately:
$$a_{1}=$$\frac{3+\sqrt{33}$$}{6}$$\approx$$$$\frac{3+5.7446}{6}$$$$\approx$$1.4574,$$ $$a_{2}=$$\frac{3-\sqrt{33}$$}{6}$$\approx$$$$\frac{3-5.7446}{6}$$$$\approx$$-0.4574.$$ Thus $$a_{1}$$\in$$(1,$$\infty$$),\qquad a_{2}$$\in$$(-$$\infty$$,0),$$ exactly the allowed intervals.

Step 7 : Nature and count of the solutions
Both $$a_{1},a_{2}$$ contain $$$$\sqrt{33}$$,$$ an irrational number, so both are irrational. No integer satisfies the equation, and there are **two** (more than one) admissible values.

Therefore the true statements are:
Option C: An irrational number satisfies the equation. (In fact both solutions are irrational.)
Option D: More than one $$a$$ satisfies the equation.