If \(I = \int\displaystyle\frac{x}{x-\sqrt{x^2-4}}\,dx = \alpha x^3 + \beta(x^2-4)^{\frac{3}{2}} + \gamma\), where \(\gamma\) is constant of integration, then
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\alpha=\beta\)
Rationalising the denominator by multiplying above and below by \(x + \sqrt{x^2-4}\):
\( \displaystyle \frac{x}{x - \sqrt{x^2-4}} = \frac{x\left(x + \sqrt{x^2-4}\right)}{x^2 - (x^2 - 4)} \)
\( \displaystyle = \frac{x^2 + x\sqrt{x^2-4}}{4} \)
So
\( \displaystyle I = \frac{1}{4}\int x^2\,dx + \frac{1}{4}\int x\sqrt{x^2-4}\,dx \)
For the second integral, put \(t = x^2 - 4\), so \(dt = 2x\,dx\):
\( \displaystyle \int x\sqrt{x^2-4}\,dx = \frac{1}{2}\int t^{1/2}\,dt = \frac{1}{3}t^{3/2} \)
\( \displaystyle = \frac{1}{3}(x^2-4)^{\frac{3}{2}} \)
Hence
\( \displaystyle I = \frac{x^3}{12} + \frac{1}{12}(x^2-4)^{\frac{3}{2}} + \gamma \)
So \(\alpha = \frac{1}{12}\) and \(\beta = \frac{1}{12}\), which gives \(\alpha = \beta\).