`int (x^2 - x^-2) dx` Find the general indefinite integral.

You need to evaluate the indefinite integral, such that:

`int f(x)dx = F(x) + c`

`int (x^2 - x^(-2))dx = int (x^2)dx - int x^(-2) dx `

Evaluating each definite integral, using the formula `int x^n dx = (x^(n+1))/(n+1) + c` , yields:

`int (x^2)dx= (x^3)/3 + c`

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

Gathering the results, yields:

`int (x^2 - x^(-2))dx = (x^3)/3 - (-1/x) + c = (x^3)/3...

You need to evaluate the indefinite integral, such that:

`int f(x)dx = F(x) + c`

`int (x^2 - x^(-2))dx = int (x^2)dx - int x^(-2) dx `

Evaluating each definite integral, using the formula `int x^n dx = (x^(n+1))/(n+1) + c` , yields:

`int (x^2)dx= (x^3)/3 + c`

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

Gathering the results, yields:

`int (x^2 - x^(-2))dx = (x^3)/3 - (-1/x) + c = (x^3)/3 + 1/x + c`

Hence, evaluating the indefinite integral yields `int (x^2 - x^(-2))dx = (x^3)/3 + 1/x + c.`