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$\int_{0}^{1} \frac{2 x+3}{5 x^{2}+1} d x$


Let $I=\int_{0}^{1} \frac{2 x+3}{5 x^{2}+1} d x$

$\int \frac{2 x+3}{5 x^{2}+1} d x=\frac{1}{5} \int \frac{5(2 x+3)}{5 x^{2}+1} d x$

$=\frac{1}{5} \int \frac{10 x+15}{5 x^{2}+1} d x$

$=\frac{1}{5} \int \frac{10 x}{5 x^{2}+1} d x+3 \int \frac{1}{5 x^{2}+1} d x$

$=\frac{1}{5} \int \frac{10 x}{5 x^{2}+1} d x+3 \int \frac{1}{5\left(x^{2}+\frac{1}{5}\right)} d x$

$=\frac{1}{5} \log \left(5 x^{2}+1\right)+\frac{3}{5} \cdot \frac{1}{\frac{1}{\sqrt{5}}} \tan ^{-1} \frac{x}{\frac{1}{\sqrt{5}}}$

$=\frac{1}{5} \log \left(5 x^{2}+1\right)+\frac{3}{\sqrt{5}} \tan ^{-1}(\sqrt{5} x)$


By second fundamental theorem of calculus, we obtain


$=\left\{\frac{1}{5} \log (5+1)+\frac{3}{\sqrt{5}} \tan ^{-1}(\sqrt{5})\right\}-\left\{\frac{1}{5} \log (1)+\frac{3}{\sqrt{5}} \tan ^{-1}(0)\right\}$

$=\frac{1}{5} \log 6+\frac{3}{\sqrt{5}} \tan ^{-1} \sqrt{5}$

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