7.2 definite integral — Maths STD 12 Science — Question

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} $ 

$\left[ \because \int_{\,-a}^{\,a}{f(x)=0,\,}\text{if }f(-x)=-f(x) \right]$.