R rboatwr0 New member Joined Apr 22, 2014 Messages 5 Apr 23, 2014 #1 ∫(x^(1/2)) / (x^(1/2)-2) dx My work so far: ∫(1/(x^(1/2)-2))*(x^(1/2)) dx trying to use ∫1/u dx = ln|u|+C u= (x^(1/2)-2) du=1/2 (x^(-1/2)) dx Now I am stuck. I am not sure how to turn the x^(-1/2) to x^(1/2)
∫(x^(1/2)) / (x^(1/2)-2) dx My work so far: ∫(1/(x^(1/2)-2))*(x^(1/2)) dx trying to use ∫1/u dx = ln|u|+C u= (x^(1/2)-2) du=1/2 (x^(-1/2)) dx Now I am stuck. I am not sure how to turn the x^(-1/2) to x^(1/2)
pka Elite Member Joined Jan 29, 2005 Messages 11,978 Apr 23, 2014 #2 rboatwr0 said: ∫(x^(1/2)) / (x^(1/2)-2) dx My work so far: ∫(1/(x^(1/2)-2))*(x^(1/2)) dx trying to use ∫1/u dx = ln|u|+C u= (x^(1/2)-2) du=1/2 (x^(-1/2)) dx Click to expand... \(\displaystyle \dfrac{\sqrt{x}}{\sqrt{x}-2}=1+\dfrac{2}{\sqrt{x}-2}\)
rboatwr0 said: ∫(x^(1/2)) / (x^(1/2)-2) dx My work so far: ∫(1/(x^(1/2)-2))*(x^(1/2)) dx trying to use ∫1/u dx = ln|u|+C u= (x^(1/2)-2) du=1/2 (x^(-1/2)) dx Click to expand... \(\displaystyle \dfrac{\sqrt{x}}{\sqrt{x}-2}=1+\dfrac{2}{\sqrt{x}-2}\)
