Let \(\int\displaystyle\frac{dx}{\left(\sqrt{x} - \sqrt{x - 1}\right)^2} = \alpha\,u(x) + \beta\,v(x) + C\), where \(u(x) = x^2 - x + \left(x - \displaystyle\frac{1}{2}\right)\sqrt{x^2 - x}\) and \(v(x) = \log\left|x - \displaystyle\frac{1}{2} + \sqrt{x^2 - x}\right|\). The value of \(\alpha + \beta\) is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{3}{4}\)
Since \(\left(\sqrt{x} - \sqrt{x - 1}\right)\left(\sqrt{x} + \sqrt{x - 1}\right) = x - (x - 1) = 1\), rationalising gives
\( \displaystyle \frac{1}{\left(\sqrt{x} - \sqrt{x - 1}\right)^2} = \left(\sqrt{x} + \sqrt{x - 1}\right)^2 \)
\( = x + (x - 1) + 2\sqrt{x(x - 1)} \)
\( = (2x - 1) + 2\sqrt{x^2 - x} \)
The first part integrates directly:
\( \displaystyle \int(2x - 1)\,dx = x^2 - x \)
For the second part, complete the square:
\( \displaystyle x^2 - x = \left(x - \frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^2 \)
and use the standard integral
\( \displaystyle \int\sqrt{t^2 - a^2}\,dt = \frac{t}{2}\sqrt{t^2 - a^2} - \frac{a^2}{2}\log\left|t + \sqrt{t^2 - a^2}\right| + C \)
with \(t = x - \frac{1}{2}\), \(a = \frac{1}{2}\):
\( \displaystyle 2\int\sqrt{x^2 - x}\,dx = 2\left[\frac{x - \frac{1}{2}}{2}\sqrt{x^2 - x} - \frac{1}{8}\log\left|x - \frac{1}{2} + \sqrt{x^2 - x}\right|\right] \)
\( \displaystyle = \left(x - \frac{1}{2}\right)\sqrt{x^2 - x} - \frac{1}{4}\log\left|x - \frac{1}{2} + \sqrt{x^2 - x}\right| \)
Adding the two parts:
\( \displaystyle \int\frac{dx}{\left(\sqrt{x} - \sqrt{x - 1}\right)^2} = \left[x^2 - x + \left(x - \frac{1}{2}\right)\sqrt{x^2 - x}\right] - \frac{1}{4}\,v(x) + C \)
\( \displaystyle = 1 \cdot u(x) - \frac{1}{4}\,v(x) + C \)
So \(\alpha = 1\) and \(\beta = -\frac{1}{4}\):
\( \displaystyle \alpha + \beta = 1 - \frac{1}{4} = \frac{3}{4} \)