What is the integral of integral of integral of 1 ?

The integral of integral of integral of 1 is (x^3)/6+(Cx^2)/2+Cx+C

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integral of integral of integral of 1

\int \int \int 1

\frac{x ^{3}}{6} + \frac{C x ^{2}}{2} + C x + C

Solution steps

\int \int \int 1

\int 1 d x = x + C

= \int \int (x + C)

\int (x + C) d x = \frac{x ^{2}}{2} + C x + C

= \int (\frac{x ^{2}}{2} + C x + C)

\int (\frac{x ^{2}}{2} + C x + C) d x = \frac{x ^{3}}{6} + \frac{C x ^{2}}{2} + C x + C

= \frac{x ^{3}}{6} + \frac{C x ^{2}}{2} + C x + C

\int_{0}^{π} \int_{0}^{π} cos (x + y) d x d y

\int_{π}^{2 π} \int_{0}^{π} (sin (x) + cos (y)) d x d y

\int_{0}^{\frac{2}{3}} \int_{y}^{2 - 2 y} (x + 2 y) e^{(y - x)} d x d y

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

\int_{0}^{4} \int_{x}^{2} sin (y^{3}) d y d x

What is the integral of integral of integral of 1 ?
The integral of integral of integral of 1 is (x^3)/6+(Cx^2)/2+Cx+C