Evaluate the definite integral $\int_{0}^{1} \frac{2 x+3}{5 x^{2}+1} d x$.

Let $I = \int_{0}^{1} \frac{2 x+3}{5 x^{2}+1} d x$.
We can split the integral as:
$I = \int_{0}^{1} \frac{2 x}{5 x^{2}+1} d x + \int_{0}^{1} \frac{3}{5 x^{2}+1} d x$.
For the first part,let $u = 5 x^{2}+1$,then $du = 10 x d x$,so $2 x d x = \frac{1}{5} du$.
$\int \frac{2 x}{5 x^{2}+1} d x = \frac{1}{5} \ln(5 x^{2}+1)$.
For the second part,$\int \frac{3}{5 x^{2}+1} d x = 3 \int \frac{1}{5(x^{2}+\frac{1}{5})} d x = \frac{3}{5} \int \frac{1}{x^{2}+(\frac{1}{\sqrt{5}})^{2}} d x$.
Using the formula $\int \frac{1}{x^{2}+a^{2}} d x = \frac{1}{a} \tan^{-1}(\frac{x}{a})$,we get:
$\frac{3}{5} \cdot \sqrt{5} \tan^{-1}(\sqrt{5} x) = \frac{3}{\sqrt{5}} \tan^{-1}(\sqrt{5} x)$.
Thus,the antiderivative is $F(x) = \frac{1}{5} \ln(5 x^{2}+1) + \frac{3}{\sqrt{5}} \tan^{-1}(\sqrt{5} x)$.
Evaluating from $0$ to $1$:
$I = F(1) - F(0) = [\frac{1}{5} \ln(6) + \frac{3}{\sqrt{5}} \tan^{-1}(\sqrt{5})] - [\frac{1}{5} \ln(1) + \frac{3}{\sqrt{5}} \tan^{-1}(0)]$.
Since $\ln(1) = 0$ and $\tan^{-1}(0) = 0$,we get:
$I = \frac{1}{5} \ln(6) + \frac{3}{\sqrt{5}} \tan^{-1}(\sqrt{5})$.