Hi! Can someone integrate the following step by step? Thank you. =] 1/[x(L-x)]dx or simply 1/(xL-x^2)dx where L is a constant.
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Originally Posted by feiyingx Hi! Can someone integrate the following step by step? Thank you. =] 1/[x(L-x)]dx or simply 1/(xL-x^2)dx where L is a constant. Use partial fractions: $\displaystyle \int \frac{1}{x(L-x)}dx=\frac{1}{L}\left[\int \frac{1}{x}+\frac{1}{L-x}\,dx\right] $ RonL
Thanks xD.
If $\displaystyle L\not =0$ then it is a CaptainBlank said. If $\displaystyle L=0$ then, $\displaystyle \int \frac{1}{-x^2}dx=\frac{1}{x}+C$
Thanks for the help. =]
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