Mathematics

Evaluate $\int { \dfrac { x+1 }{ { x }^{ 2 }+3x+12 } } dx$

SOLUTION
$\int{\cfrac{x+1}{{x}^{2} + 3x + 12} dx}$
$= \cfrac{1}{2} \int{\cfrac{2 \left( x + 1 \right)}{{x}^{2} + 3x + 12} dx}$
$= \cfrac{1}{2} \int{\cfrac{2x + 2 + 1 - 1}{{x}^{2} + 3x + 12} dx}$
$\Rightarrow \int{\cfrac{x+1}{{x}^{2} + 3x + 12} dx} = \cfrac{1}{2} \left[ \underset{\left( u \right)}{ \int{\cfrac{2x+3}{{x}^{2} + 3x + 12} dx}} - \underset{\left( v \right)}{\int{\cfrac{1}{{x}^{2} + 3x + 12}}} \right] ..... \left( 1 \right)$
Solving for $u$ and $v$, we have
$u = \int{\cfrac{2x+3}{{x}^{2} + 3x + 12} dx}$
Let ${x}^{2} + 3 + 12 = t \Rightarrow \left( 2x + 3 \right) dx = dt$
$\therefore u = \int{\cfrac{dt}{t}} = \log{t} + C$
Substituting the value of $t$, we get
$u = \log{\left( {x}^{2} + 3x + 12 \right)} + {C}_{1}$
Now,
$v = \int{\cfrac{1}{{x}^{2} + 3x + 12} dx}$
$\Rightarrow v = \int{\cfrac{1}{{\left( x + \cfrac{3}{2} \right)}^{2} - \cfrac{9}{4} + 12} dx} = \int{\cfrac{1}{{\left( x + \cfrac{3}{2} \right)}^{2} + {\left( \cfrac{\sqrt{39}}{2} \right)}^{2}} dx}$
$\therefore v = \cfrac{2}{\sqrt{39}} \tan^{-1}{\left( \cfrac{x + \cfrac{3}{2}}{\cfrac{\sqrt{39}}{2}} \right)}$
$\Rightarrow v = \cfrac{2}{\sqrt{39}} \tan^{-1}{\left( \cfrac{2x + 3}{\sqrt{39}} \right)} + {C}_{2}$
Substituting the value of $x$ in equation $\left( 1 \right)$, we have
$\Rightarrow \int{\cfrac{x+1}{{x}^{2} + 3x + 12} dx}$ $= \cfrac{1}{2} \left[ \log{\left( {x}^{2} + 3x + 12 \right)} - \cfrac{2}{\sqrt{39}} \tan^{-1}{\left( \cfrac{2x + 3}{\sqrt{39}} \right)} + {C}_{1} - {c}_{2} \right] \\ = \cfrac{1}{2} \log{\left( {x}^{2} + 3x + 12 \right)} - \cfrac{1}{\sqrt{39}} \tan^{-1}{\left( \cfrac{2x + 3}{\sqrt{39}} \right)} + C$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
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