Find the indefinite integral of $\int_{} \frac{x}{x^2+4}dx$

Question

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.

Answer 1

Choose $u=x^2+4$ and hence $\mathrm{d}u = 2 x \: \mathrm{d}x$.

Your integral becomes $$\frac{1}{2}\int\frac{\mathrm{d}u}{u} = \frac{1}{2}\ln |u| +C= \frac{1}{2}\ln|x^2+4|+C$$

Answer 2

$$\int \frac{xdx}{x^2+4}=\frac 12 \int \frac{2xdx}{x^2+4}=\frac12 \ln(x^2+4)+C$$ where we have used that

$$\frac{d}{dx}(x^2+4)=2x$$ and $$\int \frac{dt}{t}=\ln|t|+C.$$