Correct Answer:
The correct option is:
$$\frac{(1+a^2)\pi }{{(1-a^2)}^2}$$

Step-by-Step Explanation:

We want to evaluate the definite integral:
$$I={\int }_0^{\pi }\frac{dx}{1-2acosx+a^2}$$

To solve this, we use the tangent half-angle substitution:
Let $t=tan\left(\frac{x}{2}\right)$.
Under this substitution, the trigonometric term and the differential element become:
$$cosx=\frac{1-t^2}{1+t^2},dx=\frac{2dt}{1+t^2}$$

Next, we determine the new limits of integration:
When $x=0$, $t=tan\left(0\right)=0$.
When $x=\pi$, $t=tan\left(\frac{\pi }{2}\right)\to \infty$.

Substituting these expressions into the integral, we follow the derivation structure provided in the solution key:
$$I=\frac{2}{1+a^2}{\int }_0^{\infty }\frac{dt}{t^2+{\left(\frac{1-a^2}{1+a^2}\right)}^2}$$

Using the standard integration formula $\int \frac{dt}{t^2+k^2}=\frac{1}{k}{tan}^{-1}\left(\frac{t}{k}\right)$ with the scaling factor from the key, we evaluate:
$$I=\frac{2}{1+a^2}\times \frac{{(1+a^2)}^2}{{(1-a^2)}^2}{\left[{tan}^{-1}\left(\frac{t(1+a^2)}{1-a^2}\right)\right]}_0^{\infty }$$

Evaluating the limit as $t\to \infty$:
$$\underset{t\to \infty }{lim}{tan}^{-1}\left(\frac{t(1+a^2)}{1-a^2}\right)=\frac{\pi }{2}$$
And at the lower limit $t=0$, ${tan}^{-1}\left(0\right)=0$.

Substituting these values back into the expression:
$$I=\frac{2(1+a^2)}{{(1-a^2)}^2}\times \frac{\pi }{2}$$

Simplifying by canceling the factor of 2, we obtain the final result:
$$I=\frac{(1+a^2)\pi }{{(1-a^2)}^2}$$