Question

$\underset{0}{\overset{1}{\int }}\frac{2x+3}{5x^2+1}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I={\int }_0^1\frac{2x+3}{5x^2+1}\mathit{}\mathit{d}x\mathit{.}\mathit{}\mathrm{Then}\mathit{,} \\ I={\int }_0^1\frac{2x}{5x^2+1}\mathit{}\mathit{d}x+{\int }_0^1\frac{3}{5x^2+1}\mathit{}\mathit{d}x \\ \Rightarrow I=\frac{1}{5}{\int }_0^1\frac{10x}{5x^2+1}\mathit{d}x+3{\int }_0^1\frac{1}{{\left(\sqrt{5}x\right)}^2+1^2}\mathit{d}x \\ \Rightarrow I=\frac{1}{5}{\left[\log \left(5x^2+1\right)\right]}_0^1+\frac{3}{\sqrt{5}}{\left[{\tan }^{-1}\left(\sqrt{5}x\right)\right]}_0^1 \\ \Rightarrow I=\frac{1}{5}\log 6+\frac{3}{\sqrt{5}}{\tan }^{-1}\sqrt{5} \end{gathered}$$