Question

Evaluate the difinite integral: ${\int }_0^2\frac{6x+3}{x^2+4}dx$

Answer 1

${\int }_0^2\frac{6x+3}{x^2+4}dx$
Let, $F\left(x\right)=\frac{6x+3}{x^2+4}dx$
$=\stackrel{\underset{}{\int \frac{6x}{x^2+4}}dx}{I_1}+\stackrel{\underset{}{\int \frac{3}{x^2+4}}dx}{I_2}dx$

For $I_1$ :
$I_1=\int \frac{6x}{x^2+4}dx$
Put $x^2+4=t$
Differentiating w.r.t. $x$
$\Rightarrow 2xdx=dt$
Therefore,
$\int \frac{6x}{x^2+4}dx$
$=\int \frac{3}{t}dt$
$=3\log \left|\begin{array}{c}t\end{array}\right|=\log \left|\begin{array}{c}{x}^2+4\end{array}\right|$

For $I_2$:
$I_2\int \frac{3}{x^2+4}dx$
$=3\int \frac{1}{x^2+4}dx$
$=3\int \frac{1}{x^2+2^2}dx$
$=3\times \frac{1}{2}{\tan }^{-1}\frac{x}{2}$
$=\frac{3}{2}{\tan }^{-1}\frac{x}{2}$
Therefore
$F\left(x\right)=I_1+I_2$
$F\left(x\right)=3\log \left|\begin{array}{c}{x}^2+4\end{array}\right|+\frac{3}{2}{\tan }^{-1}\frac{x}{2}$
Now,
${\int }_0^2\frac{6x+3}{x^2+4}dx=F\left(2\right)-F\left(0\right)$
$(3\log \left|\begin{array}{c}{2}^2+4\end{array}\right|+\frac{3}{2}{\tan }^{-1}\frac{2}{2})-(3\log \left|\begin{array}{c}0+4\end{array}\right|-\frac{3}{2}{\tan }^{-1}\left(\frac{0}{2}\right))$
$=3\log 8+\frac{3}{2}\times \frac{\pi }{4}-3\log 4$
$=3\log 2+\frac{3\pi }{8}$