Question

How do you integrate `int 1/x^2dx`?

Answer 1

`int(1/x^2)dx = -1/x + C`

Explanation:

Use the exponent rule `a^-n = 1/a^n`:

`=int(x^-2)dx`

Use the power rule of integration, which states that `int(x^ndx) = x^(n + 1)/(n + 1) + C`, where `{n| n != -1, n in RR}`. As Jim H mentioned, it is worth noting that when faced with `int(1/x)dx`, the integral is `ln|x| + C`.

`= -x^-1 + C`

`=-1/x + C`

Hopefully this helps!