Question

what is increase integration of DX by x minus 2 the whole square

Answer 1

$I=\int\frac{dx}{(x-2)^2}\\\\=\int(x-2)^{-2}dx\\\\\\we\:know,\\\int \: x^ndx=\frac{x^{n+1}}{n+1}+C$
$so,I=\int\:(x-2)^{-2}dx=\frac{(x-2)^{-2+1}}{-2+1}+C\\\\=\frac{(x-2)^{-1}}{-1}\\\\=\frac{-1}{(x-2)}+C$

Answer 2

ur ans is -1/(x-2)
because integration of 1/(x-a)^n is (-n+1)/(x-a)^(-n+1)