$\int_{0}^{1} \frac{2 x+3}{5 x^{2}+1} d x$
Let $I=\int_{0}^{1} \frac{2 x+3}{5 x^{2}+1} d x$
$\int \frac{2 x+3}{5 x^{2}+1} d x=\frac{1}{5} \int \frac{5(2 x+3)}{5 x^{2}+1} d x$
$=\frac{1}{5} \int \frac{10 x+15}{5 x^{2}+1} d x$
$=\frac{1}{5} \int \frac{10 x}{5 x^{2}+1} d x+3 \int \frac{1}{5 x^{2}+1} d x$
$=\frac{1}{5} \int \frac{10 x}{5 x^{2}+1} d x+3 \int \frac{1}{5\left(x^{2}+\frac{1}{5}\right)} d x$
$=\frac{1}{5} \log \left(5 x^{2}+1\right)+\frac{3}{5} \cdot \frac{1}{\frac{1}{\sqrt{5}}} \tan ^{-1} \frac{x}{\frac{1}{\sqrt{5}}}$
$=\frac{1}{5} \log \left(5 x^{2}+1\right)+\frac{3}{\sqrt{5}} \tan ^{-1}(\sqrt{5} x)$
$=\mathrm{F}(x)$
By second fundamental theorem of calculus, we obtain
$I=\mathrm{F}(1)-\mathrm{F}(0)$
$=\left\{\frac{1}{5} \log (5+1)+\frac{3}{\sqrt{5}} \tan ^{-1}(\sqrt{5})\right\}-\left\{\frac{1}{5} \log (1)+\frac{3}{\sqrt{5}} \tan ^{-1}(0)\right\}$
$=\frac{1}{5} \log 6+\frac{3}{\sqrt{5}} \tan ^{-1} \sqrt{5}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.