Q. If $\int_{}^{}{\frac{\log{(x + \sqrt{1 + x^{2}})}}{\sqrt{1 + x^{2}}}\ dx = gof(x) +}$ constant, then
A) $f(x) = \log{(x + \sqrt{x^{2} + 1}})$
B) $f(x) = \log{(x + \sqrt{x^{2} + 1})}$ and $g(x) = x^{2}$
C) $f(x) = \log{(x + \sqrt{x^{2} + 1})}$ and $g(x) = \frac{x^{2}}{2}$
D) $f(x) = \frac{x^{2}}{2}$ and $g(x) = \log{(x + \sqrt{x^{2} + 1})}$
Explanation
(c)
On putting $\log{(x + \sqrt{1 + x^{2})}} = t$
$\Rightarrow \ \frac{1}{x + \sqrt{1 + x^{2}}} \times \left( 1 + \frac{x}{\sqrt{1 + x^{2}}} \right)dx = dt$
$\Rightarrow \ \frac{dx}{\sqrt{1 + x^{2}}} = dt$
$\therefore\ \int_{}^{}{\frac{\log{(x + \sqrt{1 + x^{2})}}}{\sqrt{1 + x^{2}}}dx = \int_{}^{}{t\ dt = \frac{1}{2}t^{2} + c}}$
$= \frac{1}{2}{\log{(x + \sqrt{x^{2} + 1)}}}^{2} + c$
Thus, $f(x) = \log{(x + \sqrt{x^{2} + 1)}}$
and $g(x) = \frac{x^{2}}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.