Question

Evaluate $\int \frac{dx}{x\sqrt{x^2+4}}$

• A. $\frac{1}{2}(\log x-\log (\sqrt{x^2+4}+2))$
• B. $\frac{1}{2}(\log x+\log (\sqrt{x^2+4}+2))$
• C. $\frac{1}{4}(\log x-\log (\sqrt{x^2+4}+2))$
• D. $\frac{1}{4}(\log x+\log (\sqrt{x^2+4}+2))$

Answer 1

The correct option is A $\frac{1}{2}(\log x-\log (\sqrt{x^2+4}+2))$

Let $I=\int \frac{dx}{x\sqrt{x^2+4}}$

Put $x=2\tan t\Rightarrow dx=2{\sec }^{2}tdt$

$I=2\int \frac{cosect}{4}dt=\frac{1}{2}\int -\frac{-\cot tcosect-cosec{}^{2}t}{\cot t+cosect}dt$

Put $\cot t+cosect=u\Rightarrow (-\cot tcosect-cosec{}^{2}t)dt=du$

$I=-\frac{1}{2}\int \frac{1}{u}du=-\frac{1}{2}\log u$

$I=-\frac{1}{2}\log \left(\cot t+cosect\right)$

$I=-\frac{1}{2}\log (\frac{\sqrt{x^2+4}}{x}+\frac{2}{x})$

$=\frac{1}{2}(\log x-\log (\sqrt{x^2+4}+2))$