Q100 / 120 · NDA 2024 (Phase 2) · Mathematics
Paragraph:
Let $2 \int \frac{x^2-1}{\sqrt{x^2+1}} d x=\mathrm{U}(x) \mathrm{V}(x)-3 \ln \{\mathrm{U}(x)+\mathrm{V}(x)\}+c$
Question:
What is $\mathrm{U}(x) \mathrm{V}(x)$ equal to?
Question: What is $\mathrm{U}(x) \mathrm{V}(x)$ equal to?
- A$\sqrt{x^2+x^4}$
- B$\sqrt{x+x^3}$
- C$\frac{\sqrt{x^2+x^4}}{2}$
- Correct answerD$2 \sqrt{x^2+x^4}$
Solution
$\begin{aligned} & I=2 \int \frac{x^2-1}{\sqrt{x^2+1}} d x=2 \int\left(\frac{x^2+1}{\sqrt{x^2+1}}-\frac{2}{\sqrt{x^2+1}}\right) d x \\ & =2 \int\left(\sqrt{x^2+1}-\frac{2}{\sqrt{x^2+1}}\right) d x \\ & =2\left[\frac{x}{2} \sqrt{x^2+1}+\frac{1}{2} \ln \left(x+\sqrt{x^2+1}\right)-2 \ln \left(x+\sqrt{x^2+1}\right)+c\right. \\ & =x \sqrt{x^2+1}+\ln \left(x+\sqrt{x^2+1}\right)-4 \ln \left(x+\sqrt{x^2+1}\right) \\ & =x \sqrt{x^2+1}-3 \ln \left(x+\sqrt{x^2+1}\right)+c \\ & =U(x) \cdot V(x)-3 \ln \{U(x)+V(x)\}+c \\ & \therefore U(x)=x \text { and } V(x)=\sqrt{x^2+1}\end{aligned}$
$\therefore \mathrm{U}(\mathrm{x}) \cdot \mathrm{V}(\mathrm{x})=\mathrm{x} \sqrt{\mathrm{x}^2+1}=\sqrt{\mathrm{x}^4+\mathrm{x}^2}=\sqrt{\mathrm{x}^2+\mathrm{x}^4}$
Hence, the correct answer is option: 'D'.