integral of $\frac {\sqrt {x+4}}x$

Question

So I'm trying to compute $\int\frac {\sqrt {x+4}}xdx$ . They tell me to use a substitution to make it a rational function. I set $u=\sqrt{x+4}$ and got the integrand $\frac {2u^2}{u^2-4} du$. With polynomial long division and partial fractions, I arrived at the integral of $$\int 1+\frac {1}{u+2}-\frac {1}{u-2}du$$ giving the answer of $\sqrt {x+4} + \ln|\sqrt {x+4} -2| - \ln|\sqrt {x+4} +2| + C$, which is wrong.

Answer 1

Let $I$ be the integral

$$I=\int \frac{\sqrt{x+4}}{x}\,dx$$

Now, substitute $u=\sqrt{x+4}$ to that $x=u^2-4$ and $dx=2u\,du$. Then, $I$ becomes

$$\begin{align} I&=2\int \frac{u^2}{u^2-4}\,du\\\\ &=2\int \left(\frac{u^2-4+4}{u^2-4}\right)\,du\\\\ &=2u+2\int\left(\frac{1}{u-2}-\frac{1}{u+2}\right)\,du\\\\ &=2u+2\log\left|\frac{u-2}{u+2}\right|+C\\\\ &=2\sqrt{x+4}+2\log\left|\frac{\sqrt{x+4}-2}{\sqrt{x+4}+2}\right|+C \end{align}$$