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$\log (x + \sqrt {1 + {x^2}} ) + c$
- DNone of these
✓
Answer
Correct option: A.
$\frac{1}{2}{[\log (x + \sqrt {1 + {x^2}} )]^2} + c$
a
(a) Put $\log (x + \sqrt {1 + {x^2}} ) = t \Rightarrow \frac{1}{{\sqrt {1 + {x^2}} }}\,dx = dt,$ then
(a) Put $\log (x + \sqrt {1 + {x^2}} ) = t \Rightarrow \frac{1}{{\sqrt {1 + {x^2}} }}\,dx = dt,$ then
$\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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