\[\sqrt{1 + x^{2}} + \tan^{- 1}x + c] | MATH - INTEGRATION BY SUBSTITUTION | SolveeIt

$\sqrt{1 + x^{2}} + \tan^{- 1}x + c$

$\sqrt{1 + x^{2}} - \log\{ x + \sqrt{1 + x^{2}}\} + c$

$\sqrt{1 + x^{2}} + \log\{ x + \sqrt{1 + x^{2}}\} + c$

$\sqrt{1 + x^{2}} + \log(\sec x + \tan x) + c$

None of these

Answer

$\sqrt{1 + x^{2}} + \log\{ x + \sqrt{1 + x^{2}}\} + c$

Explanation

$\frac{1}{4}\left( x + \frac{1}{x} \right)^{4} + c$

Put $\frac{x^{4}}{4} + \frac{3x^{2}}{2} + 3\log x - \frac{1}{2x^{2}} + c$

then it reduces to

$\frac{x^{4}}{4} + \frac{3x^{2}}{2} + 3\log x + \frac{1}{x^{2}} + c$

Question: \[\sqrt{1 + x^{2}} + \tan^{- 1}x + c\]...