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Question Number 133664 by shaker last updated on 23/Feb/21

	Answered by liberty last updated on 23/Feb/21
	$partial fraction (1/((x+1)^2 (x^2 +1))) = (A/(x+1))+(B/((x+1)^2 ))+((Cx+D)/(x^2 +1)) ⇔ 1=(x+1)(x^2 +1)A+(x^2 +1)B+(x+1)^2 (Cx+D) x=0⇒1=A+B+D⇒A=−B...i x=−1⇒1=D...ii x=1⇒1=4A−2A+2(C+1) −1=2A+2C...iii x=−2⇒1=−5A−5A−2C+1 0 = −10A−2C; C=−5A...iv ⇒−1=−8A; A=(1/8); B=−(1/8) C= −(5/8) ; D= 1 I=∫[(1/(8(x+1))) −(1/(8(x+1)^2 )) +((−(5/8)x+1)/(x^2 +1)) ]dx =(1/8)ln ∣x+1∣+(1/(8(x+1)))−(5/8)∫ (x/(x^2 +1))dx +arctan (x) =((ln ∣x+1∣)/8) +(1/(8x+8))+arctan (x)−(5/(16)) ln ∣x^2 +1∣ + c$
	$partial fraction$ $\frac{1}{{(x + 1)}^{2} (x^{2} + 1)} = \frac{A}{x + 1} + \frac{B}{{(x + 1)}^{2}} + \frac{Cx + D}{x^{2} + 1}$ $\Leftrightarrow 1 = (x + 1) (x^{2} + 1) A + (x^{2} + 1) B + {(x + 1)}^{2} (Cx + D)$ $x = 0 \Rightarrow 1 = A + B + D \Rightarrow A = - B . . . i$ $x = - 1 \Rightarrow 1 = D . . . ii$ $x = 1 \Rightarrow 1 = 4 A - 2 A + 2 (C + 1)$ $- 1 = 2 A + 2 C . . . iii$ $x = - 2 \Rightarrow 1 = - 5 A - 5 A - 2 C + 1$ $0 = - 10 A - 2 C; C = - 5 A . . . iv$ $\Rightarrow - 1 = - 8 A; A = \frac{1}{8}; B = - \frac{1}{8}$ $C = - \frac{5}{8}; D = 1$ $I = \int [\frac{1}{8 (x + 1)} - \frac{1}{8 {(x + 1)}^{2}} + \frac{- \frac{5}{8} x + 1}{x^{2} + 1}] dx$ $= \frac{1}{8} \ln ∣ x + 1 ∣ + \frac{1}{8 (x + 1)} - \frac{5}{8} \int \frac{x}{x^{2} + 1} dx + \arctan (x)$ $= \frac{\ln ∣ x + 1 ∣}{8} + \frac{1}{8 x + 8} + \arctan (x) - \frac{5}{16} \ln ∣ x^{2} + 1 ∣ + c$
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