Question

$\int ({\frac{1-x}{x^{2}})^2 \, dx$

Answer 1

$\displaystyle\int\left(\dfrac{1-x}{x^2}\right)^2dx=\displaystyle\int\left(\dfrac{1}{x^2}-\dfrac1x\right)^2dx=\displaystyle\int\left(\dfrac{1}{x^4}-\dfrac{2}{x^3}+\dfrac{1}{x^2}\right)dx=$

$\displaystyle\int\left(x^{-4}-2x^{-3}+x^{-2})dx=\dfrac{x^{-3}}{-3}-2\cdot\dfrac{x^{-2}}{-2}+\dfrac{x^{-1}}{-1}+C=$

$=-\dfrac{1}{3x^3}+\dfrac{1}{x^2}-\dfrac1x+C$

Dupa a doua egalitate am folosit formula $(A-B)^2=A^2-2AB+B^2$

Am folosit si:

$\int x^ndx=\dfrac{x^{n+1}}{n+1}+C$

Am folosit si :

$x^{-n}=\dfrac{1}{x^n}$