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anonymous
 4 years ago
Integration question:
\[\int\limits_{}^{} {1 \over \sqrt{3xx^2}}\]
anonymous
 4 years ago
Integration question: \[\int\limits_{}^{} {1 \over \sqrt{3xx^2}}\]

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i've tried allsort of techniques not sure how to go about

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i believe it is linked to arcsin...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0just need a pointer in right direction not full soln

ash2326
 4 years ago
Best ResponseYou've already chosen the best response.2write 3xx^2 as \(0(x^23x)\) now complete the square \[(0(x^23x+9/49/4)\] \[{(3/2)}^2{(x3/2)}^2\] now substitute x3/2 as u dx = du now use the standard formula

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0First of all complete the square in the denominator as : \[\int\limits 1/\sqrt{3x  x^2} dx = \int\limits 1/\sqrt{ [9/4(x3/2)^2]} \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Now, For the integrand 1/sqrt(9/4(x3/2)^2), substitute u = x3/2 and du = dx

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0all right with the rest

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The integral now becomes 1/sqrt(9/4u^2) du. The integral of 1/sqrt(9/4u^2) is sin^(1)((2 u)/3) Now you can proceed easily!
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