Mademoiselle23
contestada

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

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

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

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

[tex]=x- \frac{1}{x} -2ln(x) +C[/tex]

Respuesta :

LammettHash
Your answer is fine. Distributing the power and expanding the numerator is probably the cleanest approach.
[tex]\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[/tex]

I was going to say that. .but I notice you edited above, and is correct :)

Otras preguntas