Question:
Write a value of $\int \frac{1+\log x}{3+x \log x} d x$
Solution:
Let, $x(\log x)=t$
Differentiating both sides with respect to $x$
$\frac{d t}{d x}=x \frac{1}{x}+\log x=1+\log x$
$\Rightarrow d t=(1+\log x) d x$
$y=\int \frac{1}{3+t} d t$
Use formula $\int \frac{1}{a+t} d t=\log (a+t)$
$y=\log (3+t)+c$
Again, put $t=x(\log x)$
$y=\log (3+x(\log x))+c$