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[math: 4x+1(x+4)2(x2+4)] \frac{4x+1}{(x+4)^2(x^2+4)} can be expressed in partial fractions as:
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Question restatement: We are asked to express the fraction (4x+1)/((x+4)^2 (x^2+4)) in partial fractions. The provided data shows an intended form: b. [Ax+4+B(x+4)2+Cx+Dx2+4] A/(x+4) + B/(x+4)^2 + (Cx+D)/(x^2+4).
First, recall the standard partial fraction decomposition approach for a rational function with a repeated linear factor and an irreducible quadratic: when the denominator has (x+4)^2 and (x^2+4), the appropriate decomposition is of the form A/(x+4) + B/(x+4)^2 + (Cx+D)/(x^2+4). The idea is to account for the double root at x = -4 with two distinct......Login to view full explanationLog in for full answers
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Question textThe function f(x)=\dfrac{x+2}{(x+1)(x-1)^2} can be resolved into partial fractions of the form:\dfrac{a}{x+1} + \dfrac{b}{x-1} + \dfrac{c}{(x-1)^2}where a,\,b and c are real number values.Use this to express \displaystyle \int f(x)\,dx in the form\displaystyle \int f(x)\,dx = \dfrac{1}{A}\ln|x+1|+\dfrac{1}{B}\ln|x-1| + \dfrac{D}{2x-2}+ C.where A,\,B and D are integer values, and C is a constant of integration.Fill in the correct values for A,\,B, and D.A = Answer 1 Question 23[input] B = Answer 2 Question 23[input] D = Answer 3 Question 23[input]
Question textThe function f(x)=\dfrac{x+2}{(x+1)(x-1)^2} can be resolved into partial fractions of the form:\dfrac{a}{x+1} + \dfrac{b}{x-1} + \dfrac{c}{(x-1)^2}where a,\,b and c are real number values.Use this to express \displaystyle \int f(x)\,dx in the form\displaystyle \int f(x)\,dx = \dfrac{1}{A}\ln|x+1|+\dfrac{1}{B}\ln|x-1| + \dfrac{D}{2x-2}+ C.where A,\,B and D are integer values, and C is a constant of integration.Fill in the correct values for A,\,B, and D.A = Answer 1 Question 23[input] B = Answer 2 Question 23[input] D = Answer 3 Question 23[input]
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