1. ## more integrals

still having problems with integrals. i'm stumped on how to get the four substitutions. here's the problem

evaluate the integral.

$\displaystyle \int_1^2 \frac{(lnx)^2dx}{x^3}$

just getting $\displaystyle f\prime(x), f(x), g\prime(x) and g(x)$ would be a great help. thank you

2. Originally Posted by scottie.mcdonald
still having problems with integrals. i'm stumped on how to get the four substitutions. here's the problem

evaluate the integral.

$\displaystyle \int_1^2 \frac{(lnx)^2dx}{x^3}$

just getting $\displaystyle f\prime(x), f(x), g\prime(x) and g(x)$ would be a great help. thank you
$\displaystyle u=[\ln(x)]^2 \implies du=2\frac{\ln(x)}{x}dx$

$\displaystyle dv=\frac{1}{x^3} \implies v=-\frac{1}{2x^2}$

$\displaystyle \int_1^2 \frac{(lnx)^2dx}{x^3}=-\frac{[\ln(x)]^2}{2x^2}\bigg|_{1}^{2}+\int_{1}^{2}\frac{\ln(x)}{ x^3}dx$

To finish this you need to integrate by parts again.

Good luck.

3. Originally Posted by TheEmptySet
$\displaystyle u=[\ln(x)]^2 \implies du=2\frac{\ln(x)}{x}dx$

$\displaystyle dv=\frac{1}{x^3} \implies v=-\frac{1}{2x^2}$

$\displaystyle \int_1^2 \frac{(lnx)^2dx}{x^3}=-\frac{[\ln(x)]^2}{2x^2}\bigg|_{1}^{2}+\int_{1}^{2}\frac{\ln(x)}{ x^3}dx$
the $\displaystyle f\prime(x), f(x), g\prime(x) and g(x)$ seem correct but when you put them in the formula

$\displaystyle \int_1^2 g\prime(x)f(x) = f(x)g(x)|_1^2 - \int_1^2 f\prime(x)g(x)$, i am getting something different. the last integral of what I think is different then what you put down ie:

$\displaystyle \int_1^2 2(lnx)(\frac{1}{x})(\frac{1}{2x^3})dx = \int_1^2 \frac{2(lnx)dx}{2x^4}$

are you calculating a different way?