Respuesta :

Answer:

Explanation

Given the expression;

[tex]\frac{x+1}{x^2+4x+3}[/tex]

You are to express as partial fraction as shown;

[tex]\begin{gathered} \frac{x+1}{x^2+4x+3} \\ =\frac{x+1}{x^2+3x+x+3} \\ =\frac{x+1}{x(x+3)+1(x+3)} \\ =\frac{x+1}{(x+1)(x+3)} \end{gathered}[/tex]

Split the result;

[tex]\begin{gathered} \frac{x+1}{(x+3)(x+1)}=\frac{A}{x+3}+\frac{B}{x+1} \\ \frac{x+1}{(x+3)(x+1)}=\frac{A(x+1)+B(x+3)}{x+3(x+1)} \\ x+1\text{ = A(x+1)+B(x+3)} \\ \end{gathered}[/tex]

Get B by making x+1 = 0

x = -1

Subatitute x = -1 into the result;

-1+1 = A(-1+1)+B(-1+3)

0 = 0 + 2B

2B = 0

B = 0

Get A by making x+3 = 0

x = -3

Substitute x = -3 into the same result

-3+1 = A(-3+1)+B(-3+3)

-2 = -2A + 0

A = -2/-2

A = 1

Substituting A = 0 and B = 1 into the partial fraction

[tex]\frac{x+1}{(x+3)(x+1)}=\text{ }\frac{0}{x+1}+\frac{1}{x+3}[/tex]

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