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NDA 2024
Exam
NDA
Year
2024
Subject
Mathematics
Chapter
Indefinite Integration

NDA 2024 · Mathematics · Indefinite Integration

Paragraph: Let 2 ∫ x^2-1 x^2+1 d x=U(x) V(x)-3 \ U(x)+V(x)\ +c...

Paragraph: Let 2∫x2−1x2+1dx=U(x)V(x)−3ln⁡{U(x)+V(x)}+c2 \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 U(x)V(x)\mathrm{U}(x) \mathrm{V}(x) equal to?

Options

  1. Ax2+x4\sqrt{x^2+x^4}
  2. Bx+x3\sqrt{x+x^3}
  3. Cx2+x42\frac{\sqrt{x^2+x^4}}{2}
  4. D2x2+x42 \sqrt{x^2+x^4}

Correct answer

D. 2x2+x42 \sqrt{x^2+x^4}

Step-by-step solution

I=2∫x2−1x2+1dx=2∫(x2+1x2+1−2x2+1)dx=2∫(x2+1−2x2+1)dx=2[x2x2+1+12ln⁡(x+x2+1)−2ln⁡(x+x2+1)+c=xx2+1+ln⁡(x+x2+1)−4ln⁡(x+x2+1)=xx2+1−3ln⁡(x+x2+1)+c=U(x)⋅V(x)−3ln⁡{U(x)+V(x)}+c∴U(x)=x and  V(x)=x2+1\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{\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}$
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