Write a value


Write a value of $\int \frac{1+\log x}{3+x \log x} d x$


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 $t=x(\log x)$

$y=\log (3+x(\log x))+c$

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