Question

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

Answer 1

Let $I={\int }_0^1\frac{2x+3}{5x^2+1}dx$
$\Rightarrow \int \frac{2x+3}{5x^2+1}dx=\frac{1}{5}\int \frac{5(2x+3)}{5x^2+1}dx$
$=\frac{1}{5}\int \frac{10x+15}{5x^2+1}dx$
$=\frac{1}{5}\int \frac{10x}{5x^2+1}dx+\int \frac{1}{5x^2+1}dx$
$=\frac{1}{5}\int \frac{10x}{5x^2+1}dx+3\int \frac{1}{5(x^2+\frac{1}{5})}dx$
$=\frac{1}{5}\log (5x^2+1)+\frac{3}{5}\cdot \frac{1}{\frac{1}{\sqrt{5}}}{\tan }^{-1}\frac{x}{\frac{1}{\sqrt{5}}}$
$=\frac{1}{5}\log (5x^2+1)+\frac{3}{\sqrt{5}}{\tan }^{-1}\left(\sqrt{5}x\right)$
By second fundamental theorem of calculus, we obtain
$I=F\left(1\right)-F\left(0\right)$
$=\left\{\frac{1}{5}\log \right(5+1)+\frac{3}{\sqrt{5}}{\tan }^{-1}(\sqrt{5}\left)\right\}-\left\{\frac{1}{5}\right(1)+\frac{3}{\sqrt{5}}{\tan }^{-1}(0\left)\right\}$
$=\frac{1}{5}\log 6+\frac{3}{\sqrt{5}}{\tan }^{-1}\sqrt{5}$