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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 : 96 of 120

Marks: +1, -0

What is the value of β ?

Solution:

Concept:

Rationalise the integrand using the conjugate to simplify into a sum of square roots, then integrate using the power rule.

Explanation:

Given:

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

Simplify the integrand by multiplying numerator and denominator by the conjugate

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

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

Thus

I = \frac{1}{2}\int (\sqrt{x+1} + \sqrt{x-1})dx = \frac{1}{2}\left[\int (x+1)^{1/2}dx + \int (x-1)^{1/2}dx\right]

Apply the power rule:

\int u^{1/2}du = \frac{2}{3}u^{3/2}

. 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

Comparing with

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

, we get

\beta = \frac{1}{3}

Answer:

\beta = \frac{1}{3}

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