\( \int\displaystyle\frac{x}{x-\sqrt{x^2-1}}\,dx \) is equal to :
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \displaystyle\frac{x^3}{3}+\frac{(x^2-1)^{3/2}}{3}+c \), where C is an arbitrary constant.
The denominator is rationalised by multiplying above and below by \( x+\sqrt{x^2-1} \):
\( \displaystyle \frac{x}{x-\sqrt{x^2-1}}\cdot\frac{x+\sqrt{x^2-1}}{x+\sqrt{x^2-1}}=\frac{x\left(x+\sqrt{x^2-1}\right)}{x^2-(x^2-1)} \)
\( \displaystyle =x^2+x\sqrt{x^2-1} \)
\( \displaystyle \int\left(x^2+x\sqrt{x^2-1}\right)dx=\frac{x^3}{3}+\int x\sqrt{x^2-1}\,dx \)
For the second integral put \( t=x^2-1 \), \( x\,dx=\frac{dt}{2} \):
\( \displaystyle \int x\sqrt{x^2-1}\,dx=\frac{1}{2}\cdot\frac{t^{3/2}}{3/2} \)
\( \displaystyle =\frac{(x^2-1)^{3/2}}{3} \)
\( \displaystyle \int\frac{x}{x-\sqrt{x^2-1}}\,dx=\frac{x^3}{3}+\frac{(x^2-1)^{3/2}}{3}+c \)
Why the other options are incorrect
Option (1) \( x^3+\displaystyle\frac{(x^2-1)^{3/2}}{3}+c \), where C is an arbitrary constant. — Incorrect. \( \int x^2\,dx=\frac{x^3}{3} \), not \( x^3 \).
Option (2) \( \displaystyle\frac{x^4}{3}+\frac{(x^2-1)^{3/2}}{3}+c \), where C is an arbitrary constant. — Incorrect. \( \frac{x^4}{3} \) is not the integral of \( x^2 \); its derivative is \( \frac{4x^3}{3} \).
Option (3) \( \displaystyle\frac{x^3}{3}-\frac{(x^2-1)^{3/2}}{3}+c \), where C is an arbitrary constant. — Incorrect. The sign of the second term is wrong: after rationalising, the integrand is \( x^2+x\sqrt{x^2-1} \), with a plus sign.