The integral $\frac{1}{\pi }{\int }_0^{\infty }\frac{x^{2026}}{(1+x^{2026})(1+x^2)}dx$  evaluates to $\_\_\_\_\_\_\_\_\_\_\_\_\_$  (Round off to two decimal places)

The correct option is 0.25.

To evaluate the given integral, let us denote it by $I$:

$$I=\frac{1}{\pi }{\int }_0^{\infty }\frac{x^{2026}}{(1+x^{2026})(1+x^2)}dx$$

We can solve this using the substitution method. Let us substitute $x=\frac{1}{t}$.
Then, the differential is $dx=-\frac{1}{t^2}$ dt.

Let us determine the new limits of integration:
When $x\to 0$, we have $t\to \infty$.
When $x\to \infty$, we have $t\to 0$.

Substituting these values into the integral expression gives:

$$I=\frac{1}{\pi }{\int }_{\infty }^0\frac{{(1/t)}^{2026}}{(1+{(1/t)}^{2026})(1+{(1/t)}^2)}(-\frac{1}{t^2})dt$$

We use the negative sign of the differential to swap the limits of integration back to $0$ and $\infty$:

$$I=\frac{1}{\pi }{\int }_0^{\infty }\frac{1/t^{2026}}{\left(\frac{t^{2026}+1}{t^{2026}}\right)\left(\frac{t^2+1}{t^2}\right)}\frac{1}{t^2}dt$$

Simplifying the integrand by multiplying the numerator and denominator by $t^{2026}\cdot t^2$:

$$I=\frac{1}{\pi }{\int }_0^{\infty }\frac{1}{(1+t^{2026})(1+t^2)}dt$$

Since the variable of integration is dummy, we can replace $t$ with $x$:

$$I=\frac{1}{\pi }{\int }_0^{\infty }\frac{1}{(1+x^{2026})(1+x^2)}dx$$

Now, let us add the original representation of $I$ and this new representation of $I$:

$$2I=\frac{1}{\pi }{\int }_0^{\infty }\frac{x^{2026}}{(1+x^{2026})(1+x^2)}dx+\frac{1}{\pi }{\int }_0^{\infty }\frac{1}{(1+x^{2026})(1+x^2)}dx$$

Combining the integrals over the common denominator:

$$2I=\frac{1}{\pi }{\int }_0^{\infty }\frac{x^{2026}+1}{(1+x^{2026})(1+x^2)}dx$$

We notice that the term $1+x^{2026}$ appears in both the numerator and the denominator, so they cancel out:

$$2I=\frac{1}{\pi }{\int }_0^{\infty }\frac{1}{1+x^2}dx$$

The anti-derivative of $\frac{1}{1+x^2}$ is ${tan}^{-1}\left(x\right)$. Evaluating this from $0$ to $\infty$:

$$2I=\frac{1}{\pi }{\left[{tan}^{-1}\right(x\left)\right]}_0^{\infty }$$

$$2I=\frac{1}{\pi }(\frac{\pi }{2}-0)$$

$$2I=\frac{1}{2}$$

Solving for $I$:

$$I=\frac{1}{4}=0.25$$

Thus, the value of the integral is exactly 0.25.