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I have to find the integral of 1/square root(x-3)dx

this is what i did

let u=sqr root(x-3)

du=1/2*square root(x-3)dx

dx=2*square root(x-3)du

the integral of 1/square root(x-3)dx= 2*square root(x-3)(1/u)du

..ok now what i did is plug back u= square root of(x-3) in the denom

that way Im only left with 2du.

Then i integrated that to 2x2/2 which is equal to x^2.

Ultimatley I got the integral to equal X !!!! which cant be right!!!

But doesnt this make sense, I know i did not do anything wrong...

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# Verification neededproblem seems odd

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