Question

integral 0 to 2 dx/ x^2 + 5x + 6

Answer 1

$\large\underline{\sf{Solution-}}$

$\rm :\longmapsto\:\displaystyle\int_0^2\sf \: \dfrac{dx}{ {x}^{2} + 5x + 6}$

$\rm  \:  =  \: \:\:\displaystyle\int_0^2\sf \: \dfrac{1}{ {x}^{2} + 5x + 6} \: dx$

$\rm  \:  =  \: \:\:\displaystyle\int_0^2\sf \: \dfrac{1}{ {x}^{2} + 3x + 2x + 6} \: dx$

$\rm  \:  =  \: \:\:\displaystyle\int_0^2\sf \: \dfrac{1}{ x(x + 3) +2(x + 3)} \: dx$

$\rm  \:  =  \: \:\:\displaystyle\int_0^2\sf \: \dfrac{1}{ (x + 3)(x + 2)} \: dx$

$\rm  \:  =  \: \:\:\displaystyle\int_0^2\sf \: \dfrac{3 - 2}{ (x + 3)(x + 2)} \: dx$

$\rm  \:  =  \: \:\:\displaystyle\int_0^2\sf \: \dfrac{x + 3 - x - 2}{ (x + 3)(x + 2)} \: dx$

$\rm  \:  =  \: \:\:\displaystyle\int_0^2\sf \: \dfrac{(x + 3) - (x + 2)}{ (x + 3)(x + 2)} \: dx$

$\rm  \:  =  \: \:\:\displaystyle\int_0^2\sf \: \dfrac{1}{ x + 2} \: dx - \displaystyle\int_0^2\sf \: \dfrac{1}{ x + 3} \: dx$

$\rm  \:  =  \: \:\bigg( log(x + 2) \bigg)_0^2 - \bigg( log(x + 3) \bigg)_0^2$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \because \: \boxed{\displaystyle\int\sf \: \dfrac{dx}{x} = log(x) + c}$

$\rm  \:  =  \: \: log(4) - log(2) - log(5) + log(3)$

$\rm  \:  =  \: \:log\bigg(\dfrac{4 \times 3}{5 \times 2} \bigg)$

$\rm  \:  =  \: \:log\bigg(\dfrac{6}{5} \bigg)$

Additional Information :-

$\boxed{\displaystyle\int\sf \: \dfrac{dx}{ {x}^{2} + {a}^{2} } = \dfrac{1}{a} {tan}^{ - 1} \dfrac{x}{a} + c}$

$\boxed{\displaystyle\int\sf \: \dfrac{dx}{ {x}^{2} - {a}^{2} } = \dfrac{1}{2a} \: {log} \: \dfrac{x - a}{x + a} + c}$

$\boxed{\displaystyle\int\sf \: \dfrac{dx}{ \sqrt{ {x}^{2} - {a}^{2}}} = log |x + \sqrt{ {x}^{2} - {a}^{2} } | + c}$

$\boxed{\displaystyle\int\sf \: \dfrac{dx}{ \sqrt{ {x}^{2} + {a}^{2}}} = log |x + \sqrt{ {x}^{2} + {a}^{2} } | + c}$

$\boxed{\displaystyle\int\sf \: \dfrac{dx}{ \sqrt{ {a}^{2} - {x}^{2} } } = {sin}^{ - 1} \: \dfrac{x}{a} + c}$