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Mathematics 22 Online
OpenStudy (anonymous):

How to integrate this function?

OpenStudy (anonymous):

hartnn (hartnn):

did you try doing a substitution \(\Large u = x^2 +1.5^2\) ?

OpenStudy (anonymous):

so x / sqrt (u) , and then substitute back the equation after?

hartnn (hartnn):

sorry, what ? u = x^2 +1.5^2 du = ... ?

OpenStudy (anonymous):

2x

hartnn (hartnn):

you mean du = 2x dx

hartnn (hartnn):

and denominator = \(\sqrt u\) so whats your new integral look like ?

OpenStudy (anonymous):

yes. sqrt (u ) = u ^1/2 2/3 u ^ 3/2

hartnn (hartnn):

but x dx = du/2, right ?

OpenStudy (anonymous):

exactly

OpenStudy (anonymous):

yes, i think i get it ty

OpenStudy (anonymous):

\[Put \sqrt{x^2+(1.5)^2}=t,x^2=t^2-(1.5)^2,2 x dx=2 t dt,x dx=t dt\] \[I=\int\limits \frac{ x dx }{ \sqrt{x^2+(1.5)^2} }=\int\limits \frac{ t dt }{ t }=t+c=?\]

OpenStudy (anonymous):

i rewrite it.

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