MCQ
$
\int \frac{\sqrt{1+\log x}}{x} d x=\ldots \ldots \ldots \ldots+c
$
  • A
    $\frac{3}{2}(1+\log x)^{\frac{3}{2}}$
  • $\frac{2}{3}(1+\log x)^{\frac{3}{2}}$
  • C
    $2 \sqrt{1+\log x}$
  • D
    $1+\log x$

Answer

Correct option: B.
$\frac{2}{3}(1+\log x)^{\frac{3}{2}}$
(B) $\frac{2}{3}(1+\log x)^{\frac{3}{2}}$
$\int \frac{\sqrt{1+\log x}}{x} d x$
$=\int \sqrt{t} d t \quad(\because 1+\log x=t$ લેતાં $)$
$\because \frac{1}{x} d x=d t$
$=\frac{t^{\frac{3}{3}}}{\frac{3}{2}}+c=\frac{2}{3}(1+\log x)^{\frac{3}{2}}+c$

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