The correct option is:
$$\frac{1}{a+b}$$

Step-by-Step Explanation:

We are asked to evaluate the definite integral shown in the question image:
$$I={\int }_0^1\frac{a-b{x}^2}{{(a+b{x}^2)}^2}dx$$

To solve this integral, we can recognize that the integrand is the derivative of a simpler quotient function. Let us consider the function:
$$f\left(x\right)=\frac{x}{a+b{x}^2}$$

We can find the derivative of $f\left(x\right)$ with respect to $x$ using the quotient rule:
$$\frac{d}{dx}\left[\frac{u\left(x\right)}{v\left(x\right)}\right]=\frac{u^{\text{'}}\left(x\right)v\left(x\right)-u\left(x\right)v^{\text{'}}\left(x\right)}{{\left[v\right(x\left)\right]}^2}$$

Applying the quotient rule where $u\left(x\right)=x$ and $v\left(x\right)=a+b{x}^2$:
$$f^{\text{'}}\left(x\right)=\frac{\left(1\right)·(a+b{x}^2)-\left(x\right)·\left(2bx\right)}{{(a+b{x}^2)}^2}$$
Simplifying the numerator:
$$f^{\text{'}}\left(x\right)=\frac{a+b{x}^2-2b{x}^2}{{(a+b{x}^2)}^2}$$
$$f^{\text{'}}\left(x\right)=\frac{a-b{x}^2}{{(a+b{x}^2)}^2}$$

Since the derivative matches the integrand, the antiderivative of the function is:
$$\int \frac{a-b{x}^2}{{(a+b{x}^2)}^2}dx=\frac{x}{a+b{x}^2}+C$$

Now, we evaluate the definite integral by applying the integration limits from $0$ to $1$:
$$I={\left[\frac{x}{a+b{x}^2}\right]}_0^1$$
Substitute the upper limit $x=1$ and the lower limit $x=0$:
$$I=\left(\frac{1}{a+b{\left(1\right)}^2}\right)-\left(\frac{0}{a+b{\left(0\right)}^2}\right)$$
$$I=\frac{1}{a+b}-0$$
$$I=\frac{1}{a+b}$$

Thus, the value of the integral is $\frac{1}{a+b}$, which matches the option shown in the fourth image.