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

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

$\int_{1}^{e^{2}} \frac{{ln}^{2} x}{x} d x$ $int_1^(e^2)ln^2x/xdx$ =?

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Answer 1 · 2018-06-24T09:15:07

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

Explanation:

We want to evaluate

$I = \int_{1}^{e^{2}} \frac{{ln}^{2} (x)}{x} d x$ $I=int_1^(e^2)ln^2(x)/xdx$

Make a substitution $u = ln (x) \Rightarrow d u = \frac{1}{x} d x$ $color(blue)(u=ln(x)=>du=1/xdx$

New limits $x = 1 \Rightarrow u = 0$ $color(blue)(x=1=>u=0$ and $x = e^{2} \Rightarrow u = 2$ $color(blue)(x=e^2=>u=2$

$I = \int_{0}^{2} \frac{u^{2}}{x} \cdot x d u = \int_{0}^{2} u^{2} d u = \frac{1}{3} {[u^{3}]}_{0}^{3} = \frac{8}{3}$ $I=int_0^2 u^2/x*xdu=int_0^2 u^2du=1/3[u^3]_0^2=8/3$

Answer 2 · 2018-06-24T09:18:50

$\frac{8}{3}$ $8/3$

Explanation:

to integrate $\int_{1}^{e^{2}} \frac{{ln}^{2} x}{x} d x$ $int_1^(e^2)ln^2x/xdx$ first
substitute $u = ln x$ $u=lnx$ so $\frac{d u}{d x} = \frac{1}{x}$ $(du)/dx=1/x$ therby $d x = x d u$ $dx=xdu$ and $x = e^{u}$ $x=e^u$
plugging this in will give:
$\int_{0}^{2} \frac{u^{2}}{e^{u}} \cdot e^{u} d u$ $int_0^2u^2/e^u*e^udu$ the $e^{u}$ $e^u$ cancels out and we get:
$\int_{0}^{2} u^{2} d u$ $int_0^2u^2du$ now we can use power rule to get:
$\frac{u^{3}}{3} {∣ ∣ ∣}_{0}^{2}$ $u^3/3|_0^2$ now we can resubstitute $u = ln x$ $u=lnx$
and we will get: $\frac{{ln}^{3} x}{3} {∣ ∣ ∣}_{1}^{e^{2}}$ $ln^3x/3|_1^(e^2)$ which we can calculate as follows: $\frac{{ln}^{3} (e^{2})}{3} - \frac{{ln}^{3} (1)}{3} = \frac{8}{3} - \frac{0}{3} = \frac{8}{3}$ $ln^3(e^2)/3-ln^3(1)/3=8/3-0/3=8/3$

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

$\int_{1}^{e^{2}} \frac{{ln}^{2} x}{x} d x$ $int_1^(e^2)ln^2x/xdx$ =?

2 Answers

Explanation:

Explanation:

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