Theory of Intelligence
11TH•Mathematics•Partial Fractions•Exercise 5.2

Case II: Repeated Linear Factors

Practice set 5.2 (partial fractions with irreducible quadratic factors and repeated quadratic factors).

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Q1

Question 1

Resolve into partial fractions: 2x2+3x+3(x+1)(x2+1)\dfrac{2x^2+3x+3}{(x+1)(x^2+1)}

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Step-by-Step Solution

Question 1

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Solution

Let:

2x2+3x+3(x+1)(x2+1)=Ax+1+Bx+Cx2+1\frac{2x^2+3x+3}{(x+1)(x^2+1)}=\frac{A}{x+1}+\frac{Bx+C}{x^2+1}

Multiply by (x+1)(x2+1)(x+1)(x^2+1):

2x2+3x+3=A(x2+1)+(Bx+C)(x+1)2x^2+3x+3=A(x^2+1)+(Bx+C)(x+1)

Solve to get:

A=1,B=1,C=2A=1,\quad B=1,\quad C=2

Hence:

2x2+3x+3(x+1)(x2+1)=1x+1+x+2x2+1\boxed{\frac{2x^2+3x+3}{(x+1)(x^2+1)}=\frac{1}{x+1}+\frac{x+2}{x^2+1}}
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