# more integrals

• Feb 26th 2009, 06:54 PM
scottie.mcdonald
more integrals
still having problems with integrals. i'm stumped on how to get the four substitutions. here's the problem

evaluate the integral.

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

just getting $f\prime(x), f(x), g\prime(x) and g(x)$ would be a great help. thank you
• Feb 26th 2009, 07:08 PM
TheEmptySet
Quote:

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.

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

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

$u=[\ln(x)]^2 \implies du=2\frac{\ln(x)}{x}dx$

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

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

To finish this you need to integrate by parts again.

Good luck.
• Feb 26th 2009, 07:29 PM
scottie.mcdonald
Quote:

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

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

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

the $f\prime(x), f(x), g\prime(x) and g(x)$ seem correct but when you put them in the formula

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

$\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?