What is the indefinite integral of $((lnx)^2)/x$ ?

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Answer 1 · 2016-06-27T21:23:11

$((ln x)^3)/3+C$

Do a substitution: $u=ln x$ , $du = 1/x dx$ to get:

$\int\ ((ln(x))^2)/x\ dx=\int\ u^{2}\ du=u^{3}/3+C=((ln x)^3)/3+C$

Answer 2 · 2016-06-27T21:27:27

$=1/3 (ln x)^3 + C$

$int ((lnx)^2)/x \ dx$

another integration where knowing the commonest calculus patterns comes in very useful

so we have the pattern that $d/dx (f(x))^3 = 3 (f(x) )^2f'(x)$

and $d/dx ln (g(x)) = 1/(g(x)) g'(x)$

combining these ideas, watch what happens if we do

$d/dx (ln x)^3$

$= color{red}{3} (ln x)^2 * 1/x = color{red}{3} ((ln x)^2)/x$

so we do $color{red}{1/3} d/dx (ln x)^3 = d/dx (color{red}{1/3}(ln x)^3) = ((ln x)^2)/x$

if follows that

$int \ ((ln x)^2)/x \ dx =1/3 (ln x)^3 + C$

What is the indefinite integral of ((lnx)^2)/x ((lnx)^2)/x?