\( \int\displaystyle\frac{\sqrt{1+x^{2}}}{x^{4}}\,dx \) is equal to :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( -\displaystyle\frac{1}{3}\left(1+\frac{1}{x^{2}}\right)^{3/2}+C \) : Where C is an arbitrary constant
Taking x out of the root (for x > 0), \( \sqrt{1+x^{2}}=x\sqrt{1+\frac{1}{x^{2}}} \):
\( \displaystyle \int\frac{\sqrt{1+x^{2}}}{x^{4}}\,dx=\int\sqrt{1+\frac{1}{x^{2}}}\cdot\frac{1}{x^{3}}\,dx \)
Let \( t=1+\frac{1}{x^{2}} \), so \( dt=-\frac{2}{x^{3}}\,dx \), i.e. \( \frac{dx}{x^{3}}=-\frac{dt}{2} \):
\( \displaystyle =-\frac{1}{2}\int t^{1/2}\,dt \)
\( \displaystyle =-\frac{1}{2}\cdot\frac{2}{3}t^{3/2}+C \)
\( \displaystyle =-\frac{1}{3}\left(1+\frac{1}{x^{2}}\right)^{3/2}+C \)
Why the other options are incorrect
Option (1) \( \left(1+\displaystyle\frac{1}{x^{2}}\right)^{3/2}+C \) : Where C is an arbitrary constant — Incorrect. The factors \( -\frac{1}{2} \) (from dt) and \( \frac{2}{3} \) (from integrating \( t^{1/2} \)) are missing.
Option (3) \( \displaystyle\frac{1}{3}\left(1+\frac{1}{x^{2}}\right)^{3/2}+C \) : Where C is an arbitrary constant — Incorrect. The minus sign from \( dt=-\frac{2}{x^{3}}\,dx \) has been dropped.
Option (4) \( -\displaystyle\frac{1}{3}\left(1+\frac{1}{x^{2}}\right)^{5/2}+C \) : Where C is an arbitrary constant — Incorrect. Integrating \( t^{1/2} \) raises the power to \( \frac{3}{2} \), not \( \frac{5}{2} \).