Q. $\int_{}^{}\frac{\log\left( x + \sqrt{1 + x^{2}} \right)}{\sqrt{1 + x^{2}}}\ dx$ is equal to
A) $\left\lbrack \log\left( x + \sqrt{1 + x^{2}} \right) \right\rbrack^{2} + c$
B) $x\log\left( x + \sqrt{1 + x^{2}} \right) + c$
C) $\frac{1}{2}\log{\left( x + \sqrt{1 + x^{2}} \right) + c}$
D) $\frac{1}{2}\left\lbrack \log{(x + \sqrt{1 + x^{2}})} \right\rbrack^{2} + c$
Explanation
(d)
Put $x + \sqrt{1 + x^{2}} = t$
$\Rightarrow \left( 1 + \frac{1}{2\sqrt{1 + x^{2}}}.2x \right)dx = dt$
$\Rightarrow \frac{x + \sqrt{1 + x^{2}}}{\sqrt{1 + x^{2}}}\ dx = dt$
$\Rightarrow \frac{dx}{\sqrt{1 + x^{2}}} = \frac{dt}{t}$
$\therefore\ \int_{}^{}\frac{\log\left( x + \sqrt{1 + x^{2}} \right)}{\sqrt{1 + x^{2}}}\ dx = \int_{}^{}\frac{\log t}{t}\ dt$
$= \frac{\left( \log t \right)^{2}}{2} + c$
$= \frac{1}{2}{\lbrack\log{(x} + \sqrt{1 + x^{2}})\rbrack}^{2} + c$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.