\( \int x\sqrt{x^2-1}\,dx \) is equal to : ( where C is an arbitrary constant )
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{1}{3}(x^2-1)^{3/2}+C \)
Let \( t=x^2-1 \), so \( dt=2x\,dx \), that is, \( x\,dx=\frac{dt}{2} \).
\( \displaystyle \int x\sqrt{x^2-1}\,dx=\frac{1}{2}\int t^{1/2}\,dt \)
\( \displaystyle =\frac{1}{2}\cdot\frac{t^{3/2}}{3/2}+C \)
\( \displaystyle =\frac{1}{3}t^{3/2}+C \)
\( \displaystyle =\frac{1}{3}(x^2-1)^{3/2}+C \)
Shortcut: Differentiate the options instead of integrating. By the chain rule,
\( \displaystyle \frac{d}{dx}\left[\frac{1}{3}(x^2-1)^{3/2}\right]=\frac{1}{3}\cdot\frac{3}{2}(x^2-1)^{1/2}\cdot 2x \)
\( \displaystyle =x\sqrt{x^2-1} \)
Why the other options are incorrect
Option (1) \( \displaystyle\frac{1}{\sqrt{x^2-1}}+C \) — Incorrect. \( \frac{d}{dx}\left(\frac{1}{\sqrt{x^2-1}}\right)=\frac{-x}{(x^2-1)^{3/2}} \), not \( x\sqrt{x^2-1} \).
Option (2) \( \displaystyle\frac{2}{3}(x^2-1)^{3/2}+C \) — Incorrect. This drops the factor \( \frac{1}{2} \) from \( x\,dx=\frac{dt}{2} \); its derivative is \( 2x\sqrt{x^2-1} \).
Option (4) \( \displaystyle\frac{-1}{\sqrt{x^2-1}}+C \) — Incorrect. \( \frac{d}{dx}\left(\frac{-1}{\sqrt{x^2-1}}\right)=\frac{x}{(x^2-1)^{3/2}} \), not \( x\sqrt{x^2-1} \).