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 ∫ x 2 − 1 x 2 + 1 d x = U ( x ) V ( x ) − 3 ln { U ( x ) + V ( x ) } + c 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 2 ∫ x 2 + 1 x 2 − 1 d x = U ( x ) V ( x ) − 3 ln { U ( x ) + V ( x )} + c Question:
What is
∣ U 2 ( x ) − V 2 ( x ) ∣ \left|\mathrm{U}^2(x)-\mathrm{V}^2(x)\right| U 2 ( x ) − V 2 ( x ) equal to?
Step-by-step solution
I = 2 ∫ x 2 − 1 x 2 + 1 d x = 2 ∫ ( x 2 + 1 x 2 + 1 − 2 x 2 + 1 ) d x = 2 ∫ ( x 2 + 1 − 2 x 2 + 1 ) d x = 2 [ x 2 x 2 + 1 + 1 2 ln ( x + x 2 + 1 ) − 2 ln ( x + x 2 + 1 ) + c = x x 2 + 1 + ln ( x + x 2 + 1 ) − 4 ln ( x + x 2 + 1 ) = x x 2 + 1 − 3 ln ( x + x 2 + 1 ) + c = U ( x ) ⋅ V ( x ) − 3 ln { U ( x ) + V ( x ) } + c ∴ U ( x ) = x and V ( x ) = x 2 + 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} I = 2 ∫ x 2 + 1 x 2 − 1 d x = 2 ∫ ( x 2 + 1 x 2 + 1 − x 2 + 1 2 ) d x = 2 ∫ ( x 2 + 1 − x 2 + 1 2 ) d x = 2 [ 2 x x 2 + 1 + 2 1 ln ( x + x 2 + 1 ) − 2 ln ( x + x 2 + 1 ) + c = x x 2 + 1 + ln ( x + x 2 + 1 ) − 4 ln ( x + x 2 + 1 ) = x x 2 + 1 − 3 ln ( x + x 2 + 1 ) + c = U ( x ) ⋅ V ( x ) − 3 ln { U ( x ) + V ( x )} + c ∴ U ( x ) = x and V ( x ) = x 2 + 1 Now, ∣ U 2 ( x ) − V 2 ( x ) ∣ = ∣ x 2 − ( x 2 + 1 ) ∣ = ∣ x 2 − x 2 − 1 ∣ = ∣ − 1 ∣ = 1 \begin{aligned} & \text{ Now, }\left|\mathrm{U}^2(\mathrm{x})-\mathrm{V}^2(\mathrm{x})\right| \\ & =\left|\mathrm{x}^2-\left(\mathrm{x}^2+1\right)\right|=\left|\mathrm{x}^2-\mathrm{x}^2-1\right|=|-1|=1\end{aligned} Now, U 2 ( x ) − V 2 ( x ) = x 2 − ( x 2 + 1 ) = x 2 − x 2 − 1 = ∣ − 1∣ = 1 $
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