Trig substitution integral I am trying to find $$\int{ \frac {5x + 1}{x^2 + 4} dx}$$
The best approach would be to split up the fraction.  According to Wolfram Alpha, the answer is $\frac{5}{2}\ln\left(x^2 + 4\right) + \frac{1}{2}\displaystyle\arctan\left(\frac x2\right)$ which seems OK, but when I try the trig substitution: $x = 2\tan\theta$, I get an answer that is slightly different but not equivalent, and I've looked at this over and over and I couldn't quite figure out what I did wrong.
$$x = 2\tan\theta$$
$$dx = 2\sec^2\theta d\theta$$
$$\int{ \frac {5x + 1}{x^2 + 4}dx} = \frac{1}{4}\int{\frac{10\tan\theta + 1}{\sec^2\theta} 2\sec^2\theta \,d\theta}$$
$$ = \frac{1}{2}\int{10 \tan\theta + 1}\space d\theta$$
$$ = 5 \ln|\sec\theta| + \frac{\theta}{2} + C$$
We know $\theta = \displaystyle\arctan\left(\frac x2\right)$ and since $\tan\theta = \displaystyle\frac{x}{2}$, we can draw a triangle to see that $\sec\theta = \displaystyle\frac{\sqrt{x^2 + 4}}{2}$.
$$5 \ln|\sec\theta| + \frac{\theta}{2} = 5\ln\left({\frac{\sqrt{x^2 + 4}}{2}}\right) + \frac{1}{2}\arctan\left({\frac x2}\right) $$
$$= \frac{5}{2}\ln\left({\frac{x^2 + 4}{4}}\right) + \frac{1}{2}\arctan\left({\frac x2}\right)$$
But $$\frac{5}{2}\ln\left({\frac{x^2 + 4}{4}}\right) + \frac{1}{2}\arctan\left({\frac x2}\right) \neq \frac{5}{2}\ln\left(x^2 + 4\right) + \frac{1}{2}\arctan\left(\frac x2\right)$$
There seems to be a small difference between the answer provided by Alpha and the trig substitution method, but I cannot see where I made the mistake.


$$\frac{5}{2}\ln({\frac{x^2 + 4}{4}}) + \frac{1}{2}\arctan({x/2}) + c_1 \neq \frac{5}{2}\ln(x^2 + 4) + \frac{1}{2}\arctan(x/2) + C$$

Note that $$\frac{5}{2}\ln\left({\frac{x^2 + 4}{4}}\right) = \frac 52 \ln\left(x^2 + 4\right) - \frac 52 \left (\ln 4\right) = \frac 52 \ln(x^2 + 4) + c_2$$
So put $c_1 + c_2 = C$. Then the answers are equal. Solutions to an integral consist of a family of solutions $F(x) + C$, which differ only by a constant. That is, if $F(x) + C$ is the solution after computing an intregral, so is $F(x) + C_i$, for any constant $C_i \neq C$.  
In short, you're both correct!
