Question

Find the integral of X + 1 ÷ 4 + 5x -x2

Answer 1

Answer:

THE ANSWER IS 0

Step-by-step explanation:

THE ANSWER IS 0

Answer 2

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

$\rm :\longmapsto\:\displaystyle\int\sf \: \dfrac{x + 1}{4 + 5x - {x}^{2} } \: dx$

$\rm  \:  =  \: - \:\displaystyle\int\sf \: \dfrac{x + 1}{ {x}^{2} - 5x - 4} dx$

$\rm  \:  =  \: - \dfrac{1}{2} \:\displaystyle\int\sf \: \dfrac{2x + 2}{ {x}^{2} - 5x - 4} dx$

$\rm  \:  =  \: - \dfrac{1}{2} \:\displaystyle\int\sf \: \dfrac{2x - 5 + 5+ 2}{ {x}^{2} - 5x - 4} dx$

$\rm  \:  =  \: - \dfrac{1}{2} \:\displaystyle\int\sf \: \dfrac{2x - 5 + 7}{ {x}^{2} - 5x - 4} dx$

$\rm  \:  =  \: - \dfrac{1}{2} \:\displaystyle\int\sf \: \dfrac{2x - 5}{ {x}^{2} - 5x - 4}dx - \dfrac{1}{2} \:\displaystyle\int\sf \: \dfrac{7}{ {x}^{2} - 5x - 4}dx$

$\rm  \:  =  \: \:I_1 + I_2$

where,

$\rm :\longmapsto\:I_1 = - \dfrac{1}{2} \:\displaystyle\int\sf \: \dfrac{2x - 5}{ {x}^{2} - 5x - 4}$

and

$\rm :\longmapsto\:I_2 = - \dfrac{1}{2} \:\displaystyle\int\sf \: \dfrac{7}{ {x}^{2} - 5x - 4}$

$\rm :\longmapsto\:\displaystyle\int\sf \: \dfrac{x + 1}{4 + 5x - {x}^{2} } \: dx = I_1 + I_2 - - - (1)$

Now,

Consider,

$\rm :\longmapsto\:I_1 = - \dfrac{1}{2} \:\displaystyle\int\sf \: \dfrac{2x - 5}{ {x}^{2} - 5x - 4} dx$

we know,

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

$\rm :\longmapsto\:2x - 5 = \dfrac{d}{dx} ( {x}^{2} - 5x - 4)$

So, given integral, can be rewritten as

$\rm :\longmapsto\:I_1 = - \dfrac{1}{2} \:\displaystyle\int\sf \: \dfrac{ \dfrac{d}{dx} ( {x}^{2} - 5x - 4)}{ {x}^{2} - 5x - 4} dx$

$\rm :\implies\:I_1 = - \dfrac{1}{2}log | {x}^{2} - 5x - 4 | - - - (2)$

Now,

Consider,

$\rm :\longmapsto\:I_2 = - \dfrac{1}{2} \:\displaystyle\int\sf \: \dfrac{7}{ {x}^{2} - 5x - 4} dx$

$\rm  \:  =  \: \: - \dfrac{7}{2} \:\displaystyle\int\sf \: \dfrac{1}{ {x}^{2} - 5x - 4} dx$

$\rm  \:  =  \: \: - \dfrac{7}{2} \:\displaystyle\int\sf \: \dfrac{1}{ {x}^{2} - 5x + {\bigg(\dfrac{5}{2} \bigg) }^{2} - {\bigg(\dfrac{5}{2} \bigg) }^{2} - 4} dx$

$\rm  \:  =  \: \: - \dfrac{7}{2} \:\displaystyle\int\sf \: \dfrac{1}{ {\bigg(x - \dfrac{5}{2} \bigg) }^{2} - {\bigg(\dfrac{25}{4} \bigg) } - 4} dx$

$\rm  \:  =  \: \: - \dfrac{7}{2} \:\displaystyle\int\sf \: \dfrac{1}{ {\bigg(x - \dfrac{5}{2} \bigg) }^{2} - {\bigg(\dfrac{25 + 16}{4} \bigg) } } dx$

$\rm  \:  =  \: \: - \dfrac{7}{2} \:\displaystyle\int\sf \: \dfrac{1}{ {\bigg(x - \dfrac{5}{2} \bigg) }^{2} - {\bigg(\dfrac{41}{4} \bigg) } } dx$

$\rm  \:  =  \: \: - \dfrac{7}{2} \:\displaystyle\int\sf \: \dfrac{1}{ {\bigg(x - \dfrac{5}{2} \bigg) }^{2} - {\bigg(\dfrac{ \sqrt{41} }{2} \bigg) } ^{2} } dx$

$\rm  \:  =  \: \: - \dfrac{7}{2} \times \dfrac{1}{2 \times \dfrac{ \sqrt{41} }{2} } log \bigg |\dfrac{x - \dfrac{5}{2} - \dfrac{ \sqrt{41} }{2} }{x - \dfrac{5}{2} + \dfrac{ \sqrt{41} }{2} } \bigg|$

$\rm  \:  =  \: \: - \dfrac{7}{2 \sqrt{41} } log \bigg |\dfrac{2x -5 - \sqrt{41} }{2x - 5 + \sqrt{41} } \bigg| - - - (3)$

Now, on substituting the values from equation (2) and (3) in equation (1), we get

$\rm :\longmapsto\:\displaystyle\int\sf \: \dfrac{x + 1}{4 + 5x - {x}^{2} } \: dx = I_1 + I_2$

$\rm  \:  =- \dfrac{1}{2}log | {x}^{2} - 5x - 4 |\: - \dfrac{7}{2 \sqrt{41} } log \bigg |\dfrac{2x -5 - \sqrt{41} }{2x - 5 + \sqrt{41} } \bigg| + c$

Additional Information :-

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

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

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

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

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

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

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