Question

Evaluate : ${\int }_0^1\frac{2x+3}{5x^2+1}dx$

Answer 1

$I={\int }_0^1\frac{2x+3}{5x^2+1}dx={\int }_0^1\frac{2x}{5x^2+1}dx+{\int }_0^1\frac{3}{5x^2+1}dx$

$\Rightarrow I=I_1+I_2$

$I_1={\int }_0^1\frac{2x}{5x^2+1}dx$

Put $5x^2+1=t\Rightarrow 5\times 2xdx=dt$ and limits will be

Lower limit $x=0\Rightarrow t=1$ and Upper limit $x=1\Rightarrow t=6$

$I_1=\frac{1}{5}{\int }_1^6\frac{1}{t}dt=\frac{1}{5}|\ln |t|{|}_1^6=\frac{1}{5}|\ln 6-\ln 1|=\frac{1}{5}\ln 6$

$I_2={\int }_0^1\frac{3}{5x^2+1}dx=\frac{3}{5}{\int }_0^1\frac{1}{x^2+\frac{1}{5}}dx$

Integral of ${\int }_a^b\frac{1}{x^2+a^2}dx={\left|\frac{1}{a}{\tan }^{-1}\frac{x}{a}\right|}_a^b$

Here, $a=\frac{1}{\sqrt{5}}$

$I_2=\frac{3}{5}{\left|\sqrt{5}{\tan }^{-1}\right(\sqrt{5}x\left)\right|}_0^1=\frac{3}{\sqrt{5}}{\tan }^{-1}\left(\sqrt{5}\right)$

$\therefore I=\frac{1}{5}\ln 6+\frac{3}{\sqrt{5}}{\tan }^{-1}\left(\sqrt{5}\right)$