Is there a mistake in my evaluation of $\int\frac{3x+1}{\sqrt{5x^2+1}}dx$? I want to evaluate the following integral: 

$$\int\frac{3x+1}{\sqrt{5x^2+1}}dx$$

The answer given in my textbook is $\frac{3}{5}\sqrt{5x^2+1}+\frac{1}{\sqrt{5}}\ln(x\sqrt{5}+\sqrt{5x^2+1})$.
I think the author has excess factor $\sqrt{5}$ in logarithm.
My solution is
$$\begin{align}
\int\frac{3x+1}{\sqrt{5x^2+1}}dx &= \frac{3}{2}\int\frac{d(x^2)}{\sqrt{5x^2+1}}+\int\frac{1}{\sqrt{5x^2+1}}dx \\[6pt]
&=\frac{3}{\sqrt{5}}\sqrt{x^2+\frac{1}{5}}+\frac{1}{\sqrt{5}}\ln\left(x+\sqrt{x^2+\frac{1}{5}}\right)
\end{align}$$
since 
$$\int\frac{dx}{\sqrt{x^2+a}}=\ln\left(x+\sqrt{x^2+a}\right)$$
Am I right?
N.B. Just to be short I skipped additive constant in antiderivatives, so don't mention it.

hint for another approach
Put $$x=\frac {\sinh (t)}{\sqrt {5}} $$
then it becomes
$$\frac {1}{\sqrt {5}}\int \frac { \frac {3\sinh (t)}{\sqrt {5}}+1}{\cosh (t)}\cosh (t)dt $$
$$=\frac {3}{5}\cosh (t)+\frac {t}{\sqrt {5}}+C $$
