Integrals — Class 12 Maths Solution

: Let $I = \int\limits_0^2 {\cfrac{{6x + 3}}{{{x^2} + 4}}dx} = \int\limits_0^2 {\cfrac{{6x}}{{{x^2} + 4}}dx} + \int\limits_0^2 {\cfrac{3}{{{x^2} + 4}}dx}$

$= 3\int\limits_0^2 {\cfrac{{2x}}{{{x^2} + 4}}dx + \left[ {3 \times \cfrac{1}{2}{{\tan }^{ - 1}}\cfrac{x}{2}} \right]} _0^2$

Let ${I_1} = 3\int\limits_0^2 {\cfrac{{2x}}{{{x^2} + 4}}dx}$

Put${x^2} + 4 = t$ $\Rightarrow$ $2x\,dx = dt$
When $x = 0,t = 4$ and when $x = 2,t = 8$

$\therefore$ ${I_1} = 3\int\limits_4^8 {\cfrac{{dt}}{t} = \left[ {3\log t} \right]_4^8 = 3\left( {\log 8 - \log 4} \right) = 3\log 2}$

$\Rightarrow$ $I = 3\log 2 + \cfrac{3}{2}\left[ {{{\tan }^{ - 1}}1 - {{\tan }^{ - 1}}0} \right]$

$= 3\log 2 + \cfrac{3}{2} \times \cfrac{\pi }{4} = 3\log 2 + \cfrac{{3\pi }}{8}$