Question

\[\int\limits_{1}^{3} (x/(x^2+3))\]

Answer 1

use substiution
\[u =x^2+3\] then \[du =2x dx\] =>\[(1/2)du=xdx\]
changing bounds
upper \[u=3^3+3=12\]lower, \[u=1^2+3=4\]
now we have
\[\int\limits_{4}^{12} ((1/2)u)du\]
\[=(1/2)\int\limits\limits_{4}^{12} (u)du\]
\[=(1/2)[ u^2/2]|_{4}^{12}\]
\[=(1/2)[ 12^2/2 - 4^2/2]\]
\[=32\]

Answer 2

\[\int\limits_{1}^3{\frac{x}{x^2+3}}dx\]
\[\text{Let }u=x^2+3\].....

Answer 3

are you sure? when I do it on my calculator i get the answer (ln(3))/2 as a answer. it is the same answer i get from using wolframalpha too.

Answer 4

my problem is that i have know idea how to set it up manually

Answer 5

you might be right 1 min

Answer 6

sorry I set up the intergal worng it should be this \[\int\limits_{4}^{12} (1/2)(1/u)du\]
\[ =(1/2)\int\limits\limits_{4}^{12}(1/u)du\]
\[ =(1/2)[\ln u]|_{4}^{12}\]
\[ =(1/2)[\ln 12 -\ln4]\]
\[ =(\ln 3)/2\]

Answer 7

no problem. Thanks :)