If \(\int\displaystyle\frac{dx}{x-\sqrt{x^2-1}}= \alpha x^2+\beta x\sqrt{x^2-1}+\gamma\log\left|x+\sqrt{x^2-1}\right|+C\), then the value of α + β + γ is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{1}{2}\)
Rationalise the denominator by multiplying the numerator and the denominator by \(x+\sqrt{x^2-1}\):
\( \displaystyle \frac{1}{x-\sqrt{x^2-1}} = \frac{x+\sqrt{x^2-1}}{x^2-(x^2-1)} \)
\( = x+\sqrt{x^2-1} \)
Using the standard integral \(\int\sqrt{x^2-a^2}\,dx = \frac{x}{2}\sqrt{x^2-a^2} - \frac{a^2}{2}\log\left|x+\sqrt{x^2-a^2}\right| + C\) with a = 1:
\( \displaystyle \int\left(x+\sqrt{x^2-1}\right)dx = \frac{x^2}{2} + \frac{x}{2}\sqrt{x^2-1} - \frac{1}{2}\log\left|x+\sqrt{x^2-1}\right| + C \)
Comparing:
\( \displaystyle \alpha = \frac{1}{2},\quad \beta = \frac{1}{2},\quad \gamma = -\frac{1}{2} \)
\( \displaystyle \alpha + \beta + \gamma = \frac{1}{2} + \frac{1}{2} - \frac{1}{2} = \frac{1}{2} \)