Theory of Intelligence
11TH•Mathematics•Partial Fractions•Question 1

Solution

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

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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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