Question

Prove that
${\int }_0^1\frac{2n+3}{5n^2+1}dx$

Answer 1

${\int }_0^1\frac{2n+3}{5n^2+1}dn$

$=\frac{1}{5}{\int }_0^1\frac{10n}{5n^2+1}dn+\frac{1}{5}\int \frac{3}{n^2+\frac{1}{5}}dn$

$5n^2+1=t$

10n dn = dt

$=\frac{1}{5}{\int }_0^1\frac{1}{t}dt+\frac{3}{5}\int \frac{1}{n^2+{\left(\frac{1}{\sqrt{5}}\right)}^2}dn$

$=\frac{1}{5}{\left[log\right(5n^2+1\left)\right]}_0^1+\frac{3}{5}\left(\sqrt{5}\right){\left[ta{n}^{-1}n\sqrt{5}\right]}_0^1$

$=\frac{log6}{5}+\frac{3}{\sqrt{5}}ta{n}^{-1}\sqrt{5}$