If $I = \int \frac{dx}{(x-2)\left(1+\sqrt{7x-10-x^2}\right)} = f(t) + c$ (Where $t = \sqrt{\frac{5-x}{x-2}}$ and $f(0) = k\ln\frac{3-\sqrt{5}}{3+\sqrt{5}}$ $(k > 0)$ then $k^2$ is equal to ____.
Step-by-Step Solution
Key Concept: Use the substitution $\sqrt{7x-10-x^2} = (x-2)t$ to convert the quadratic radical into a rational function amenable to partial fractions.
Let $\sqrt{7x-10-x^2} = (x-2)t$, then squaring gives $(5-x)(x-2) = (x-2)^2t^2$, so $x-2 = \frac{3}{t^2+1}$. Substituting yields $1+\sqrt{7x-10-x^2} = \frac{t^2+3t+1}{t^2+1}$. With $x = \frac{2t^2+5}{t^2+1}$, we get $dx = \frac{-6t}{(t^2+1)^2}dt$. The integral becomes $I = -\ln|t^2+3t+1| + \frac{3}{\sqrt{5}}\ln\left|\frac{2t+3-\sqrt{5}}{2t+3+\sqrt{5}}\right| + c$ after partial fractions decomposition of the resulting rational expression.
Correct Answer: 1