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…

Question

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`

Answer 1

`\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 :)

Answer 2

Your answer is fine. Distributing the power and expanding the numerator is probably the cleanest approach.