What is the integral of 1/(1+(x^2)^{22)}x^2)^2 ?

The integral of 1/(1+(x^2)^{22)}x^2)^2 is (x^{16})/(x^{176)+1}+C

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integral of 1/(1+(x^2)^{22)}x^2)^2

\int \frac{1}{1 + ( ( ( x ^{2} ) ^{2} ) ^{2} ) ^{22}} d (((x^{2})^{2})^{2})^{2}

\frac{x ^{16}}{x ^{176} + 1} + C

Solution steps

\int \frac{1}{1 + ( ( ( x ^{2} ) ^{2} ) ^{2} ) ^{22}} d (((x^{2})^{2})^{2})^{2}

Integral of a constant:

\int a d x = a x

= \frac{1}{( ( ( x ^{2} ) ^{2} ) ^{2} ) ^{22} + 1} (((x^{2})^{2})^{2})^{2}

Simplify

\frac{1}{( ( ( x ^{2} ) ^{2} ) ^{2} ) ^{22} + 1} (((x^{2})^{2})^{2})^{2} : \frac{x ^{16}}{x ^{176} + 1}

= \frac{x ^{16}}{x ^{176} + 1}

Add a constant to the solution

= \frac{x ^{16}}{x ^{176} + 1} + C

What is the integral of 1/(1+(x^2)^{22)}x^2)^2 ?
The integral of 1/(1+(x^2)^{22)}x^2)^2 is (x^{16})/(x^{176)+1}+C