integrate $\int(\ln{x})^2dx$
Solution:
To integrate this function, we use integration by parts(IBP) which is
$\int{u}dv=uv-\int{v}du$
let $u=(\ln{x})^2$ and $dv=dx$ so that we can integrate dv to get v.
$\int{dv}=\int{dx}\Rightarrow{v}=x$ and $du=\frac{2\ln{x}}{x}dx$.
substitute into the integration by parts formula
$=x(\ln{x})^2-2\int\ln{x}dx.............(1)$
we carry out integration by parts again in eqn(1)
let $u=\ln{x}$ and $\int{dv}=\int{dx}$
$du=\frac{1}{x}dx$ and $v=x$
substitute $du$ and $v$ into the integration by parts formula yields
$=x\ln{x}-\int{x}.\frac{1}{x}dx$
$=x\ln{x}-\int{1}dx$
$=x\ln{x}-x$...................(2)
substitute (2) into the integral part of (1)
$=x(\ln{x})^2-2[x\ln{x}-x]+c$
where c is the constant of integration.
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