What is the integral of integral of 1/(x^2) ?

The integral of integral of 1/(x^2) is -ln(x)+Cx+C

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integral of integral of 1/(x^2)

\int \int \frac{1}{x ^{2}} d x d x

- ln (x) + C x + C

Solution steps

\int \int \frac{1}{x ^{2}} d x d x

\int \frac{1}{x ^{2}} d x = - \frac{1}{x} + C

= \int (- \frac{1}{x} + C) d x

\int (- \frac{1}{x} + C) d x = - ln (x) + C x + C

= - ln (x) + C x + C

\int \int cos (x + y) d x d y

\int \int sin (x + y) d x d y

\int \int (x + 2 y) d A

\int_{0}^{2} \int_{0}^{1} s + t d s d t

\int \int (x^{2} - y) d x d y

What is the integral of integral of 1/(x^2) ?
The integral of integral of 1/(x^2) is -ln(x)+Cx+C