The discussion focuses on solving the improper integral using integration techniques. The user initially applied u-substitution to reach the integral limit of \(\lim_{R\rightarrow 3-} \int_{-3}^{R-3} \left(\frac{u + 3}{\sqrt{u}}\right) du\). However, another participant suggests that integration by parts is unnecessary and recommends rearranging the integrand into simpler components: \(\frac{u + 3}{u^{1/2}} = u^{1/2} + 3u^{-1/2}\). This approach simplifies the integration process significantly.
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ggcheck said:I used u substitution to get to this point:
\lim_{R\rightarrow\ 3-}} \int_{-3}^{R-3} (\frac{u + 3}{sqrt{u}}) du
is the only way to proceed from here using integration by parts?
ahhh, I didn't think to break up the fraction, for some reason I missed thatsutupidmath said:If your calculations are correct up to this point, i do not think you need to use integ by parts. Simply just try to rearrange the integrand like this
(u+3)/u^1/2= u/u^1/2 +3/u^1/2 = u^1-1/2 +3u^-1/2 and i guess you will be fine!