MCQ
$\int_{}^{} {\frac{{\log (x + \sqrt {1 + {x^2}} )}}{{\sqrt {1 + {x^2}} }}\;dx = } $
- ✓$\frac{1}{2}{[\log (x + \sqrt {1 + {x^2}} )]^2} + c$
- B$\log {(x + \sqrt {1 + {x^2}} )^2} + c$
- C$\log (x + \sqrt {1 + {x^2}} ) + c$
- Dએકપણ નહિ.
$\int_{}^{} {\frac{{\log (x + \sqrt {1 + {x^2}} )}}{{\sqrt {1 + {x^2}} }}\,dx} = \int_{}^{} {t\,dt} $
$\int_{}^{} {\frac{{{t^2}}}{2}dt} $ $ = \frac{1}{2}{\left[ {\log (x + \sqrt {1 + {x^2}} )} \right]^2} + c$.
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