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

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

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integral of 1/((x^2)^2)x^2)^2

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

1 + C

Solution steps

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

(((x^{2})^{2})^{2})^{2} = x^{16}

= \int \frac{1}{x ^{16}} d (((x^{2})^{2})^{2})^{2}

Integral of a constant:

\int a d x = a x

= \frac{1}{x ^{16}} (((x^{2})^{2})^{2})^{2}

Simplify

\frac{1}{x ^{16}} (((x^{2})^{2})^{2})^{2} : 1

= 1

Add a constant to the solution

= 1 + C

\frac{d}{d x} (e^{2 x^{6}} - 1)

\frac{\partial}{\partial y} (x^{2} + y^{2} + x y)

\frac{d}{d x} (\frac{10}{( 1 - x ) ^{3}})

\int (- \frac{6}{x ^{4}}) d x

d er i v a t i v e 2^{t}

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