What is the value of \alpha ?

Consider the following for the next two (02) items that follow : Let \int\frac{dx}{\sqrt{x+1}-\sqrt{x-1}}=\alpha(x+1)^{\frac{3}{2}}+\beta(x-1)^{\frac{3}{2}}+c

  1. A. \frac{1}{3}
  2. B. \frac{2}{3}
  3. C. 1
  4. D. \frac{4}{3}

Correct Answer: A. \frac{1}{3}

Explanation

Rationalize the denominator by multiplying the numerator and denominator by \sqrt{x+1}+\sqrt{x-1}. The integral becomes \int \frac{\sqrt{x+1}+\sqrt{x-1}}{(x+1)-(x-1)}\,dx = \frac{1}{2}\int ( (x+1)^{1/2} + (x-1)^{1/2} )\,dx. Evaluating this gives \frac{1}{2} \left[ \frac{2}{3}(x+1)^{3/2} + \frac{2}{3}(x-1)^{3/2} \right] + c = \frac{1}{3}(x+1)^{3/2} + \frac{1}{3}(x-1)^{3/2} + c. Comparing this with the given expression, \alpha = \frac{1}{3}.

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