Question

evaulate integral x^2+x+1/(x+1)^2(x+2)*dx​

Answer 1

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \huge \red{AηsωeR } ✍$

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$\sf \:  ⟼ \int\dfrac{ {x}^{2} + x + 1 }{ {(x + 1)}^{2} \: (x + 2) } dx$

$\large\underline\red{\bold{❥︎Step :- 1 }}$

☆Using concept of Partial Fraction

$\sf \:  \dfrac{ {x}^{2} + x + 1 }{ {(x + 1)}^{2} \: (x + 2) } = \dfrac{A}{x + 1} + \dfrac{B}{ {(x + 1)}^{2} } + \dfrac{C}{x + 2} \sf \:  ⟼ \: (1)$

☆On taking LCM, we get

$\sf \:  {x}^{2} + x + 1 = A(x + 1)(x + 2) + B(x + 2) + C {(x + 1)}^{2} \sf \:  ⟼(2)$

$\large\underline\red{\bold{❥︎Step :- 2 }}$

$\bf \:  ⟼ Substituting \: 'x = - 1' \: in \: (2) , we \: get$

$\sf \:  ⟼1 - 1 + 1 = B$

$\bf\implies \:B \: = \: 1 \: \bf \:  ⟼ \: (3)$

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$\large\underline\red{\bold{❥︎Step :- 3 }}$

$\bf \:  ⟼ Substituting 'x = - 2' \: in \: (2), we \: get$

$\sf \:  ⟼ {( - 2)}^{2} - 2 + 1 = C {( - 2 + 1)}^{2}$

$\sf \:  ⟼4 - 2 + 1 = C$

$\bf\implies \:C = 3\sf \:  ⟼ \: (4)$

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$\large\underline\red{\bold{❥︎Step :- 4 }}$

$\bf \:  ⟼ Substituting 'x = 0' \: in \:(2), \: we \: get$

$\sf \:  0 + 0 + 1 = 2A + 2B + C$

☆On substituting the values of B and C, we get

$\sf \:  ⟼1 = 2A + 2 + 3$

$\sf \:  ⟼2A = - 4$

$\bf\implies \:A \: = \: - 2 \: \bf \:  ⟼ \: (5)$

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$\large\underline\red{\bold{❥︎Step :- 5 }}$

☆On substituting the values of A, B and C evaluated in equation (3), (4) and (5) in equation (1), we get

$\sf \:  \dfrac{ {x}^{2} + x + 1 }{ {(x + 1)}^{2} \: (x + 2) } = \dfrac{ - 2}{x + 1} + \dfrac{1}{ {(x + 1)}^{2} } + \dfrac{3}{x + 2}$

$\sf \:  \int\dfrac{ {x}^{2} + x + 1 }{ {(x + 1)}^{2} \: (x + 2) }dx = \int\dfrac{ - 2}{x + 1}dx + \int\dfrac{1}{ {(x + 1)}^{2} }dx + \int\dfrac{3}{x + 2}dx$

$\sf \:  \int\dfrac{ {x}^{2} + x + 1 }{ {(x + 1)}^{2} \: (x + 2) }dx = - 2 log(x + 1) - (x + 1) + 3 log(x + 2) + c$

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$\bf \:  ⟼ \int\dfrac{1}{x} dx = log(x) + c \\ \bf \:  ⟼ \int {x}^{n} dx = \dfrac{ {x}^{n + 1} }{n + 1} + c$

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