A calculus problem by Rohith M.Athreya

Relevant wiki: Integration Tricks

given integral is $\displaystyle \large \int_{e}^{e^2} (lnt)^2 dt$

let $e^x=t$ now it becomes $\displaystyle \large \int_{1}^{2} (x^2) e^x dx$

also we have $\displaystyle \large \int e^x(f(x)+f'(x))dx=e^x(f(x))+C$

the given integral can be manipulated as

$\displaystyle \large \int_{1}^{2} (x^2) e^x dx= \int_{1}^{2} (x^2+2x) e^x dx- \int_{1}^{2} (2x+2) e^x dx+ \int_{1}^{2} (2) e^x dx$

by the above property this simplifies to

$\displaystyle \large \left((x^2-2x+2)e^x\right)_1^2=2e^2-e$

Relevant wiki: Differentiation Under the Integral Sign

$\begin{aligned} I & = \int_e^{e^2} \ln^2 t \ dt & \small \color{#3D99F6} \text{Let } t = e^x, \ dt = e^x \ dx \\ & = \int_1^2 x^2 e^x \ dx \\ & = \int_1^2 \frac {\partial^2 e^{ax}}{\partial a^2} \ dx & \small \color{#3D99F6} \text{where }a = 1 \\ & = \frac {\partial^2}{\partial a^2} \int_1^2 e^{ax} \ dx \\ & = \frac {\partial^2}{\partial a^2} \left[ \frac {e^{ax}}a \right]_1^2 \\ & = \frac {\partial}{\partial a} \left[ \frac {xe^{ax}}a - \frac {e^{ax}}{a^2} \right]_1^2 \\ & = \frac {x^2e^{ax}}a - \frac {xe^{ax}}{a^2} - \frac {x^2e^{ax}}{a^2} + \frac {2xe^{ax}}{a^3} \bigg|_1^2 & \small \color{#3D99F6} \text{Putting back }a = 1 \\ & = x^2e^x - xe^x - x^2e^x + 2xe^x \bigg|_1^2 \\ & = xe^x \bigg|_1^2 \\ & = 2e^2 - e \approx \boxed{12.06} \end{aligned}$

Use integration by parts twice to get the answer:

$\int_{e}^{e^2} (\ln t)^2 dt = \int_{e}^{e^2} 1 (\ln t)^2 dt$

Let $u = (\ln t)^2$ and $dv=dt$

Then $du = \frac{2 \ln t}{t}$ and $v = t$

Hence, $\int_{e}^{e^2} (\ln t)^2 dt = \Big[t (\ln t)^2 \Big]_{e}^{e^2} - \int_{e}^{e^2} 2 \ln t dt$

$= 4e^2 - e - 2 \int_{e}^{e^2} \ln t dt$

Integrate $\int_{e}^{e^2} \ln t dt$ by parts again:

Let $u = \ln t$ and $dv = dt$

Then $du = \frac{1}{t} dt$ and $v = t$

Hence, $\int_{e}^{e^2} \ln t dt = \Big[t \ln t \Big]_{e}^{e^2} - \int_{e}^{e^2} dt = 2 e^2 - e - e^2 + e = e^2$

Hence again, $\int_{e}^{e^2} (\ln t)^2 dt$

$= 4e^2 - e - 2(e^2)$

$=2e^2 - e$

$= \boxed{12.06}$ to 2 decimal places.

A Pretty simple one .

Integrate by parts using ln^2(t) as first function and 1 as second one.

And again integrate by parts using ln(t) as first function and 1 as second one

to get e(2e-1)