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How can I solve this integral :
\int\limits ({ (1-x)/(x) })^2 \, dx

Is what I'm doing correct? Or am I just making it a lot more complicated than it is?


\int\limits ({ (1-x)/(x) })^2 \, dx = \int\limits ({ (1)/(x) (1-x)} )^2 \, dx = \int\limits ({ (1)/(x^2) (1-2x+x^2)} ) \, dx


= \int\limits ({ (1)/(x^2) - (2)/(x) + 1} ) \, dx


=x- (1)/(x) -2ln(x) +C

User Chrystie
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2 Answers

4 votes

\bf \displaystyle \int \left( \cfrac{1-x}{x} \right)^2 dx\implies \int \left(\cfrac{1}{x^2}-\cfrac{2}{x}+1 \right)dx\implies -\cfrac{1}{x}-2ln|x|+x+C

I was going to say that. .but I notice you edited above, and is correct :)
User Yagger
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8.5k points
6 votes
Your answer is fine. Distributing the power and expanding the numerator is probably the cleanest approach.
User Nikola Schou
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