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UPSC NDA I 2024 Mathematics Paper

Question Numbers: 95-96

Consider the following for the items that follow:

Let

\int \frac{d x}{\sqrt{x+1}-\sqrt{x-1}}=\alpha(x+1)^{\frac{3}{2}}+ \beta(x-1)^{\frac{3}{2}}+c

Question : 95 of 120

Marks: +1, -0

What is the value of α ?

Solution:

Concept:

Integration by rationalization and the power rule:

\int x^n \, dx = \frac{x^{n+1}}{n+1} + C

Explanation:

We rationalize the denominator by multiplying numerator and denominator by

\sqrt{x+1} + \sqrt{x-1}

The denominator becomes

(\sqrt{x+1})^2 - (\sqrt{x-1})^2 = (x+1) - (x-1) = 2

Thus,

I = \int \frac{dx}{\sqrt{x+1} - \sqrt{x-1}} = \frac{1}{2} \int (\sqrt{x+1} + \sqrt{x-1}) \, dx

Integrate each term using the power rule:

\int \sqrt{x+1} \, dx = \frac{2}{3}(x+1)^{3/2}

, and similarly for the second term.

So,

I = \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

Compare with given expression:

\alpha (x+1)^{3/2} + \beta (x-1)^{3/2} + C

Hence,

\alpha = \frac{1}{3}

Answer:

\alpha = \frac{1}{3}

(Option A).

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