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Mathematics 13 Online
OpenStudy (anonymous):

Evaluate the integral Integral (2x-1)^(1/2)/(2x+3) Please just say how to start it off.

OpenStudy (anonymous):

\[\int\limits_{}^{}\frac{\sqrt{2x-1}}{2x+3}dx\]

OpenStudy (anonymous):

That is the integral, but how do you start solving it?

OpenStudy (anonymous):

ok, one approach is to set u=sqrt(2x-1) so du=1/sqrt(2x-1)dx and dx=sqrt(2x-1)du or just dx=u du This gives:\[\int\limits_{}^{}\frac{u^2}{u^2+4} du\]

OpenStudy (anonymous):

That gets rid of you peskey radical

OpenStudy (anonymous):

So, do you get \[\sqrt{2x-1}-2\tan^{-1} (\sqrt{2x-1}/2) +C\]

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