Question
$\int_{}^{} {\frac{1}{{x\sqrt {1 + \log x} }}\;dx = } $
अंतः $\int_{}^{} {\frac{{dx}}{{x\sqrt {1 + \log x} }}} = \int_{}^{} {\frac{{dt}}{{{t^{1/2}}}} = 2{t^{1/2}} + c} = 2{(1 + \log x)^{1/2}} + c$.
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