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
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