Evaluate the following integrals:


Evaluate: $\int \frac{(1+\log x)^{2}}{x} d x$.


let $1+\log x=t$

Differentiating on both sides we get,

$\frac{1}{x} d x=d t$

Substituting it in $\int \frac{(1+\log x)^{2}}{x}$ we get,

$=\int t^{2} d t$


$=\frac{(1+\log x)^{3}}{3}+c$

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