Question

${\int }_0^2\frac{5x+1}{x^2+4}dx$

Answer 1

We have,

${\int }_0^2\frac{5x+1}{x^2+4}dx$

$I=\frac{5}{2}{\int }_0^2\frac{2x}{x^2+4}dx+{\int }_0^2\frac{dx}{x^2+4}$

Let

$x^2+4=t2xdx=dtt=4t=8$

$\int \frac{dx}{a^2+x^2}=\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)+C$

Therefore,

$\Rightarrow \frac{5}{2}{\int }_0^2\frac{dt}{t}+{\int }_0^2\frac{dx}{{\left(2\right)}^2+x^2}\Rightarrow \frac{5}{2}{\left[\log t\right]}_4^8+{\left[\frac{1}{2}{\tan }^{-1}\left(\frac{x}{2}\right)\right]}_0^2\Rightarrow \frac{5}{2}\left[\log 8-\log 4\right]+\left[\frac{1}{2}{\tan }^{-1}\left(1\right)-\frac{1}{2}{\tan }^{-1}\left(0\right)\right]\Rightarrow \frac{5}{2}\log \left(\frac{8}{4}\right)+\frac{1}{2}\left(\frac{\pi }{4}\right)-\frac{1}{2}\left(0\right)\Rightarrow \frac{5}{2}\log 2+\frac{\pi }{8}$

Hence, this is the answer.