(ln^2 x)/x integral from 1 to e^2 is?

${\int}_{1}^{{e}^{2}} {\ln}^{2} \frac{x}{x} \mathrm{dx}$ =?

Jun 24, 2018

$I = \frac{8}{3}$

Explanation:

We want to evaluate

$I = {\int}_{1}^{{e}^{2}} {\ln}^{2} \frac{x}{x} \mathrm{dx}$

Make a substitution color(blue)(u=ln(x)=>du=1/xdx

New limits color(blue)(x=1=>u=0 and color(blue)(x=e^2=>u=2

$I = {\int}_{0}^{2} {u}^{2} / x \cdot x \mathrm{du} = {\int}_{0}^{2} {u}^{2} \mathrm{du} = \frac{1}{3} {\left[{u}^{3}\right]}_{0}^{2} = \frac{8}{3}$

Jun 24, 2018

$\frac{8}{3}$

Explanation:

to integrate ${\int}_{1}^{{e}^{2}} {\ln}^{2} \frac{x}{x} \mathrm{dx}$ first
substitute $u = \ln x$ so $\frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{x}$ therby $\mathrm{dx} = x \mathrm{du}$ and $x = {e}^{u}$
plugging this in will give:
${\int}_{0}^{2} {u}^{2} / {e}^{u} \cdot {e}^{u} \mathrm{du}$ the ${e}^{u}$ cancels out and we get:
${\int}_{0}^{2} {u}^{2} \mathrm{du}$ now we can use power rule to get:
${u}^{3} / 3 {|}_{0}^{2}$ now we can resubstitute $u = \ln x$
and we will get: ${\ln}^{3} \frac{x}{3} {|}_{1}^{{e}^{2}}$ which we can calculate as follows: ${\ln}^{3} \frac{{e}^{2}}{3} - {\ln}^{3} \frac{1}{3} = \frac{8}{3} - \frac{0}{3} = \frac{8}{3}$