Q246 / 270 · NDA 2024 (Phase 1) · Mathematics
Let $\int \frac{d x}{\sqrt{x+1}-\sqrt{x-1}}=\alpha(x+1)^{\frac{3}{2}}+\beta(x-1)^{\frac{3}{2}}+c$
What is the value of $\beta$ ?
- A$-\frac{2}{3}$
- B$-\frac{1}{3}$
- Correct answerC$\frac{1}{3}$
- D$\frac{2}{3}$
Solution
Given,
$\int \frac{d x}{\sqrt{x+1}-\sqrt{x-1}}=\alpha(x+1)^{\frac{3}{2}}+\beta(x-1)^{\frac{3}{2}}+c$ $\qquad ....(i)$
Let, $\quad \mathrm{I}=\int \frac{d x}{\sqrt{x+1}-\sqrt{x-1}}$
$\begin{aligned}
& =\int \frac{(\sqrt{x+1}+\sqrt{x-1}) d x}{(\sqrt{x+1}-\sqrt{x-1})(\sqrt{x+1}+\sqrt{x-1})} \\
& \Rightarrow \quad \mathrm{I}=\frac{1}{2} \int \sqrt{(x+1)} d x+\frac{1}{2} \int \sqrt{x-1} d x \\
& \Rightarrow \quad \mathrm{I}=\frac{1}{2} \cdot \frac{(x+1)^{\frac{3}{2}}}{\left(\frac{3}{2}\right)}+\frac{\frac{1}{2}(x-1)^{\frac{3}{2}}}{\left(\frac{3}{2}\right)}+c \\
& \Rightarrow \quad \mathrm{I}=\frac{1}{3}(x+1)^{\frac{3}{2}}+\frac{1}{3}(x-1)^{\frac{3}{2}}+c \\
& \Rightarrow \frac{d x}{\sqrt{x+1}-\sqrt{x-1}}=\frac{1}{3}(x+1)^{\frac{3}{2}}+\frac{1}{3}(x-1)^{\frac{3}{2}}+c \ldots(ii)
\end{aligned}$
From (i) and (ii) we get
$\beta=\frac{1}{3}$
Hence, the correct answer is option: 'C'.