# Thread: Integral [1 to 2] (lnx)^2/(x^3) Please help

1. ## Integral [1 to 2] (lnx)^2/(x^3) Please help

Need help solving integral [1 to 2] (lnx)^2/(x^3) Please im really stuck

2. let $\displaystyle u=\ln(x) \iff e^{u}=x \iff e^{2u}=x^2 \iff e^{-2u}=\frac{1}{x^2}$
and
$\displaystyle du=\frac{1}{x}dx$ this gives

$\displaystyle \displaystyle \int_{1}^{2}(\ln(x))^2dx=\int_{0}^{\ln(2)}u^2e^{-2u}du$

Now just integrate by parts twice.

3. Originally Posted by TheEmptySet
let $\displaystyle u=\ln(x) \iff e^{u}=x \iff e^{2u}=x^2 \iff e^{-2u}=\frac{1}{x^2}$
and
$\displaystyle du=\frac{1}{x}dx$ this gives

$\displaystyle \displaystyle \int_{1}^{2}(\ln(x))^2dx=\int_{0}^{\ln(2)}u^2e^{-2u}du$

Now just integrate by parts twice.
So:
∫ x^2 * e^(2x) dx = uv - ∫ v du
= x^2*e^(2x)/2 - ∫ e^(2x)/2 (2x dx)
= x^2*e^(2x)/2 - ∫ x*e^(2x) dx

Letting:

u = x, du = dx
dv = e^(2x), v = e^(2x)/2 dx

∫ x^2 * e^(2x) dx = x^2*e^(2x)/2 - (uv - ∫ v du)
= x^2*e^(2x)/2 - [x*e^(2x)/2 - ∫ e^(2x)/2 dx]
= x^2*e^(2x)/2 - x*e^(2x)/2 + 1/2 ∫ e^(2x) dx
= x^2*e^(2x)/2 - x*e^(2x)/2 + e^(2x)/4 + C
= e^(2x)/4 * (2x^2 - 2x + 1) + C ?? Is this correct

4. Originally Posted by TheEmptySet
let $\displaystyle u=\ln(x) \iff e^{u}=x \iff e^{2u}=x^2 \iff e^{-2u}=\frac{1}{x^2}$
and
$\displaystyle du=\frac{1}{x}dx$ this gives

$\displaystyle \displaystyle \int_{1}^{2}(\ln(x))^2dx=\int_{0}^{\ln(2)}u^2e^{-2u}du$

Now just integrate by parts twice.
Or you could just do parts twice without any substitution.

5. Have a look at this website.
Be sure to click "show steps".

6. Originally Posted by kensington
Need help solving integral [1 to 2] (lnx)^2/(x^3) Please im really stuck
If the function to be integrated is...

$\displaystyle \displaystyle f(x)= \frac{\ln^{2} x}{x^{3}}$ (1)

... then integrating by parts three times You obtain...

$\displaystyle \displaystyle \int \frac{\ln^{2} x}{x^{3}}\ dx = - \frac{\ln^{2} x}{2 x^{2}} - \frac{\ln x}{2 x^{2}} - \frac{1}{4 x^{2}} + c$ (2)

... so that is...

$\displaystyle \displaystyle \int_{1}^{2} \frac{\ln^{2} x}{x^{3}}\ dx = - \frac {\ln^{2} 2 + \ln 2}{8} - \frac{1}{16} + \frac{1}{4} = .040799756902...$ (3)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

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