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Question Number 43027 by Raj Singh last updated on 06/Sep/18

Commented by math khazana by abdo last updated on 07/Sep/18

$l e t I = \int \frac{x^{2} + 1}{{(x + 1)}^{3} (x - 2)} d x l e t d e c o m p o s e$ $F (x) = \frac{x^{2} + 1}{{(x + 1)}^{3} (x - 2)}$ $F (x) = \frac{a}{x + 1} + \frac{b}{{(x + 1)}^{2}} + \frac{c}{{(x + 1)}^{3}} + \frac{d}{x - 2}$ $c = {l i m}_{x \to - 1} {(x + 1)}^{3} F (x) = - \frac{2}{3}$ $d = {l i m}_{x \to 2} (x - 2) F (x) = \frac{5}{27}$ ${l i m}_{x \to + \infty} x F (x) = 0 = a + d \Rightarrow a = - \frac{5}{27} \Rightarrow$ $F (x) = \frac{- 5}{27 (x + 1)} + \frac{b}{{(x + 1)}^{2}} - \frac{2}{3 {(x + 1)}^{3}} + \frac{5}{27 (x - 2)}$ $F (0) = - \frac{1}{2} = \frac{- 5}{27} + b - \frac{2}{3} - \frac{5}{54} \Rightarrow$ $- 1 = - \frac{10}{27} + 2 b - \frac{4}{3} - \frac{5}{27} = - \frac{15}{27} + 2 b - \frac{4}{3}$ $= - \frac{5}{9} - \frac{4}{3} + 2 b = - \frac{15 + 36}{27} + 2 b = - \frac{51}{27} + 2 b \Rightarrow$ $2 b = \frac{51}{27} - 1 = \frac{51 - 27}{27} = \frac{24}{27} = \frac{8}{9} \Rightarrow b = \frac{4}{9} \Rightarrow$ $F (x) = \frac{- 5}{27 (x + 1)} + \frac{4}{9 {(x + 1)}^{2}} - \frac{2}{3 {(x + 1)}^{3}} + \frac{5}{27 (x - 2)}$ $I = \int F (x) d x = \frac{- 5}{27} l n ∣ x + 1 ∣ - \frac{4}{9 (x + 1)} + \frac{1}{3 {(x + 1)}^{2}}$ $+ \frac{5}{27} l n ∣ x - 2 ∣ + c .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 06/Sep/18

	$2) \int {l o g}_{10} x d x$ $= \int \frac{{l o g}_{e} x}{{l o g}_{e} 10} d x$ $= \frac{1}{l n 10} \int l n x d x$ $= \frac{1}{l n 10} [l n x \times x - \int \frac{1}{x} \times x d x]$ $= \frac{1}{l n 10} [x l n x - x] + c$
	Answered by tanmay.chaudhury50@gmail.com last updated on 06/Sep/18
	$1)∫((x^2 +1)/((x+1)^3 (x−2)))dx ((x^2 +1)/((x+1)^3 (x−2)))=(a/(x+1))+(b/((x+1)^2 ))+(c/((x+1)^3 ))+(d/((x−2))) ((x^2 +1)/((x+1)^3 (x−2)))=((a(x+1)^2 (x−2)+b(x+1)(x−2)+c(x−2)+d(x+1)^3 )/((x+1)^3 (x−2))) x^2 +1=a(x+1)^2 (x−2)+b(x+1)(x−2)+c(x−2)+d(x+1)^3 a(x^2 +2x+1)(x−2)+b(x^2 −2x+x−2)+c(x−2)+d(x^3 +3x^2 +3x+1) a(x^3 +2x^2 +x−2x^2 −4x−2)+b(x^2 −x−2)+c(x−2)+d(x^3 +3x^2 +3x+1) a(x^3 −3x−2)+b(x^2 −x−2)+c(x−2)+d(x^3 +3x^2 +3x+1) x^3 (a+d)+x^2 (b+3d)+x(−3a−b+c+3d)+(−2a−2b−2c+d) now a+d=0 b+3d=1 (−3a−b+c+3d)=0 (−2a−2b−2c+d)=1 putting the value of a and b in terms of d in 3rd 4th eqn {−3(−d)−(1−3d)+c+3d}=0 3d−1+3d+c+3d=0 c=1−9d {−2(−d)−2(1−3d)−2(1−9d)+d=1 2d−2+6d−2+18d+d=1 27d=5 d=(5/(27)) a=((−5)/(27)) b=1−3((5/(27)))=1−(5/9)=(4/9) c=1−9((5/(27)))=1−(5/3)=((−2)/3) so the intregation is ∫(a/(x+1))dx+∫(b/((x+1)^2 ))dx+∫(c/((x+1)^3 ))dx+∫(d/((x−2)))dx =((−5)/(27))ln(x+1)+(4/9)×(1/((x+1)×−1))+((−2)/3)×(1/((x+1)^2 ×−2))+(5/(27))ln(x−2)+c =((−5)/(27))ln(x+1)−(4/9)×(1/(x+1))+(1/3)×(1/((x+1)^2 ))+(5/(27))ln(x−2)+c$
	$1) \int \frac{x^{2} + 1}{{(x + 1)}^{3} (x - 2)} d x$ $\frac{x^{2} + 1}{{(x + 1)}^{3} (x - 2)} = \frac{a}{x + 1} + \frac{b}{{(x + 1)}^{2}} + \frac{c}{{(x + 1)}^{3}} + \frac{d}{(x - 2)}$ $\frac{x^{2} + 1}{{(x + 1)}^{3} (x - 2)} = \frac{a {(x + 1)}^{2} (x - 2) + b (x + 1) (x - 2) + c (x - 2) + d {(x + 1)}^{3}}{{(x + 1)}^{3} (x - 2)}$ $x^{2} + 1 = a {(x + 1)}^{2} (x - 2) + b (x + 1) (x - 2) + c (x - 2) + d {(x + 1)}^{3}$ $a (x^{2} + 2 x + 1) (x - 2) + b (x^{2} - 2 x + x - 2) + c (x - 2) + d (x^{3} + 3 x^{2} + 3 x + 1)$ $a (x^{3} + 2 x^{2} + x - 2 x^{2} - 4 x - 2) + b (x^{2} - x - 2) + c (x - 2) + d (x^{3} + 3 x^{2} + 3 x + 1)$ $a (x^{3} - 3 x - 2) + b (x^{2} - x - 2) + c (x - 2) + d (x^{3} + 3 x^{2} + 3 x + 1)$ $x^{3} (a + d) + x^{2} (b + 3 d) + x (- 3 a - b + c + 3 d) + (- 2 a - 2 b - 2 c + d)$ $n o w$ $a + d = 0$ $b + 3 d = 1$ $(- 3 a - b + c + 3 d) = 0$ $(- 2 a - 2 b - 2 c + d) = 1$ $p u t t i n g t h e v a l u e o f a a n d b i n t e r m s o f d i n 3 r d$ $4 t h e q n$ ${- 3 (- d) - (1 - 3 d) + c + 3 d} = 0$ $3 d - 1 + 3 d + c + 3 d = 0$ $c = 1 - 9 d$ ${- 2 (- d) - 2 (1 - 3 d) - 2 (1 - 9 d) + d = 1$ $2 d - 2 + 6 d - 2 + 18 d + d = 1$ $27 d = 5 d = \frac{5}{27}$ $a = \frac{- 5}{27}$ $b = 1 - 3 (\frac{5}{27}) = 1 - \frac{5}{9} = \frac{4}{9}$ $c = 1 - 9 (\frac{5}{27}) = 1 - \frac{5}{3} = \frac{- 2}{3}$ $s o t h e i n t r e g a t i o n i s$ $\int \frac{a}{x + 1} d x + \int \frac{b}{{(x + 1)}^{2}} d x + \int \frac{c}{{(x + 1)}^{3}} d x + \int \frac{d}{(x - 2)} d x$ $= \frac{- 5}{27} l n (x + 1) + \frac{4}{9} \times \frac{1}{(x + 1) \times - 1} + \frac{- 2}{3} \times \frac{1}{{(x + 1)}^{2} \times - 2} + \frac{5}{27} l n (x - 2) + c$ $= \frac{- 5}{27} l n (x + 1) - \frac{4}{9} \times \frac{1}{x + 1} + \frac{1}{3} \times \frac{1}{{(x + 1)}^{2}} + \frac{5}{27} l n (x - 2) + c$
	Commented by math khazana by abdo last updated on 07/Sep/18

	$y o u r a n s w e r i s c o r r e c t s i r T a n m a y .$
	Commented by tanmay.chaudhury50@gmail.com last updated on 07/Sep/18

	$t h a n k y o u s i r . . .$
	Commented by malwaan last updated on 07/Sep/18

	$thank you$
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