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Another one is x/(2x-1)^1/2 with an upper interval of 5 and the lower being 1
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try letting u=2x-1 so du=2dx
since you have x on top, you will also need to put that in terms of u. since u=2x-1, then x=(u+1)/2. The upper limit is 2(5)-1 and the lower limit is 2(1)-1
\[1/4\int\limits_{1}^{9}(u+1)/(u ^{1/2}) du\]
\[1/4 \int\limits_{1}^{9}(u ^{1/2}+u^{-1/2} )du\]
Where are getting 1/4 from?
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