Find the indefinite integral of $\int_{} \frac{x}{x^2+4}dx$ I am beginning to question whether the indefinite integral actually exists or I am doing something wrong with my u-substitution.
Let $u = x^2 + 4, du = 2xdx,$
$$
\begin{align}
\int_{} \frac{x}{x^2+4}dx 
&= \int_{}x(x^2 + 4)^{-1} \\
&= \frac{1}{2} \int_{} u^{-1}du \\
&= \frac{1}{2} \frac{u^0}{0} = ???
\end{align}
$$
Then I tried to choose a different $u$
Let $u=x^2, du = 2xdx$
$$\int_{}\frac{x}{x^2+4} = \int_{} \frac{\sqrt u}{u^2 + 4} = ... $$
But I still run into the same problem trying to use the power rule to simplify the integral. I am beginning to think that an indefinite integral actually does not exist, but I am not sure what basis I have to assert that statement.

$$\int \frac{xdx}{x^2+4}=\frac 12 \int \frac{2xdx}{x^2+4}=\frac12 \ln(x^2+4)+C$$ where we have used that
$$\frac{d}{dx}(x^2+4)=2x$$ and $$\int \frac{dt}{t}=\ln|t|+C.$$
