Question

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

Answer 1

Let I=$∫^1_0 \frac{2x+3}{5x^2+1}dx$

$∫\frac{2x+3}{5x^2+1}dx=\frac{1}{5}∫5\frac{2x+3}{5x^2+1}dx$

$=\frac{1}{5}∫\frac{10x+15}{5x^2+1}dx$

$=\frac{1}{5}∫\frac{10x}{5x^2+1}dx+3\int\frac{1}{5x^2+1}$

$=\frac{1}{5}log(5x^2+1)+\frac{3}{5}.\frac{1}{\frac{1}{\sqrt5}} tan^{-1}\frac{x}{\frac{1}{√5}}$

$=\frac{1}{5}log(5x^2+1)+\frac{3}{√5}tan^{-1}(√5x)$

$=F(x)$

By second fundamental theorem of calculus,we obtain

$I=F(1)-F(0)$

={\frac{1}{5}log(5+1)+\frac{3}{√5}tan^{-1}(√5)}$$-\frac{1}{5}log(1)+\frac{3}{√5}tan^{-1}(0)

$=\frac{1}{5}log6+\frac{3}{√5}tan^{-1}√5$