If $\int_0^{\ln t} \frac{t \ln 2}{x^2+t^2} dt = \frac{\pi \ln 2}{4}$ $(x>0)$ then the number of integral values of $'x'$ satisfying this equation is____.
Step-by-Step Solution
Key Concept: The integral ∫₀^(ln t) (t·ln 2)/(x² + t²) dt evaluates using the arctangent antiderivative formula ∫du/(a² + u²) = (1/a)arctan(u/a). Setting this equal to π·ln2/4 yields arctan(ln t/x) = π/4, which implies ln t/x = 1, giving t = e^x. Finding integral values of x requires solving the implicit equation carefully.
Given $I = \int_0^{\pi/2} \frac{\ln(\cot\theta)\cdot\csc^2\theta}{x^2(1+\tan^2\theta)} d\theta$, substitute $d\theta = \frac{1}{x}\int(\ln x + \ln\tan\theta)d\theta$ to obtain $I = \frac{\ln x}{x}\int_0^{\pi/2} d\theta + \frac{1}{x}\int_0^{\pi/2} \ln\tan\theta\, d\theta$. Since $\int_0^{\pi/2} \ln\tan\theta\, d\theta = 0$, we have $I = \frac{\pi\ln x}{2x}$. Setting $\frac{\pi\ln x}{2x} = \frac{\pi\ln 2}{4}$ gives $\ln x = \frac{\ln 2}{2}$, so $x = 2$ or $x = 4$ (from the graph of $y = \frac{\ln x}{x}$, which has two values for most $y$-values in $(0, 1/e)$).
Correct Answer: 2