You've a typo in the denominator of your distribution function: $x$ should be $x^2$, w/o the square of the parantheses. And, as @whuber said, you need a constant term after the integration (for $x\geq 0$):

$$F(x)=\int \frac{2xdx}{(1+x^2)^2}=-\frac{1}{1+x^2}+C$$

By definition, $\lim_{x\rightarrow \infty}F(x)=1$, substituting yields $$\lim_{x\rightarrow\infty}F(x)=C-\lim_{x\rightarrow\infty}\frac{1}{1+x^2}=C=1$$

So, you can plot the following, for $x\geq 0$, (for $x<0$, $F(x)=0$): $$F(x)=1-\frac{1}{1+x^2}=\frac{x^2}{1+x^2}$$

You've a typo in the denominator of your distribution function: $x$ should be $x^2$, w/o the square of the parentheses. And, as @whuber said, you need a constant term after the integration (for $x\geq 0$):

$$F(x)=\int \frac{2xdx}{(1+x^2)^2}=-\frac{1}{1+x^2}+C$$

By definition, $\lim_{x\rightarrow \infty}F(x)=1$, substituting yields $$\lim_{x\rightarrow\infty}F(x)=C-\lim_{x\rightarrow\infty}\frac{1}{1+x^2}=C=1$$

So, you can plot the following, for $x\geq 0$, (for $x<0$, $F(x)=0$): $$F(x)=1-\frac{1}{1+x^2}=\frac{x^2}{1+x^2}$$