MCQ
$\int_{\, - \,1}^{\,1} {\log (x + \sqrt {{x^2} + 1} )\,dx = } $
- ✓$0$
- B$log\, 2$
- C$\log \frac{1}{2}$
- DNone of these
Now,$f( - x) = \log \left( {\sqrt {1 + {x^2}} - x} \right) = \log (\sqrt {1 + {x^2}} - x).\frac{{(\sqrt {1 + {x^2}} + x)}}{{(\sqrt {1 + {x^2}} + x)}}$
$ = \log \frac{{[(1 + {x^2}) - {x^2}]}}{{(\sqrt {1 + {x^2}} + x)}}$
$ = \log 1 - \log (\sqrt {1 + {x^2}} + x)$
$ = - \log (\sqrt {1 + {x^2}} + x)$
$ = - f(x)$
Hence, $\int_{\, - 1}^{\,1} {\log \,(x + \sqrt {1 + {x^2}} ) = 0} $
$\left[ \because \int_{\,-a}^{\,a}{f(x)=0,\,}\text{if }f(-x)=-f(x) \right]$.
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