Answer
$$\int\frac{x}{x^2+4}dx=\frac{1}{2}\ln(x^2+4)+c$$
Work Step by Step
To evaluate the integral
$$\int\frac{x}{x^2+4}dx$$
we will use substitution $x^2+4=t$ which gives us $2xdx=dt\Rightarrow xdx=\frac{dt}{2}$. Putting this into the integral we get:
$$\int\frac{x}{x^2+4}dx=\int\frac{1}{t}\frac{dt}{2}=\frac{1}{2}\int\frac{1}{t}dt=\frac{1}{2}\ln|t|+c$$
where $c$ is arbitrary constant. Now we have to express solution in terms of $x$:
$$\int\frac{x}{x^2+4}dx=\frac{1}{2}\ln|t|+c=\frac{1}{2}\ln(x^2+4)+c$$