Question:
Mark the correct alternative in each of the following:
The value of $\int \frac{1}{\mathrm{x}+\mathrm{x} \log \mathrm{x}} \mathrm{dx}$ is
A. $1+\log x$
B. $x+\log x$
C. $x \log (1+\log x)$
D. $\log (1+\log x)$
Solution:
$1=\int \frac{1}{x\left(1+\log _{e} x\right)} d x$
$\Rightarrow \operatorname{let}\left(1+\log _{\mathrm{e}} \mathrm{x}\right)=\mathrm{t}\left[\frac{d t}{d x}=\frac{1}{x}\right]$
$\Rightarrow \int \frac{1}{t} d t=\log _{e} t$
$\Rightarrow 1=\log (1+\log x)+C$
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