Answer
\[\frac{3x+4}{(x-3)(x^2+4x+5)}=\frac{1}{2(x-3)}-\frac{x+1}{2(x^2+4x+5)}\]
Work Step by Step
Because $x^2+4x+5$ has no real zero therefore
we can not factor it so
In this case the general form of the partial fraction decomposition is
\[\frac{3x+4}{(x-3)(x^2+4x+5)}=\frac{A}{(x-3)}+\frac{Bx+C}{x^2+4x+5}\;\;\;...(*)\]
\[3x+4=A(x^2+4x+5)+(Bx+C)(x-3)\]
\[3x+4=A(x^2+4x+5)+(Bx^2-3Bx+Cx-3C)\]
\[3x+4=(A+B)x^2+(4A-3B+C)x+(5A-3C)\]
Compare like coefficients both sides
\[A+B=0\Rightarrow B=-A\;\;\;...(1)\]
\[4A-3B+C=3\;\;\;...(2)\]
\[5A-3C=4\;\;\;...(3)\]
Using (1) in (2) then (2) becomes
\[7A+C=3\;\;\;...(4)\]
Multiply (4) by 3 then add to (3)
\[26A=13\Rightarrow A=\frac{1}{2}\]
From (1)
\[B=\frac{-1}{2}\]
From (4)
\[C=\frac{-1}{2}\]
From (*)
\[\frac{3x+4}{(x-3)(x^2+4x+5)}=\frac{1}{2(x-3)}+\frac{\frac{-1}{2}x+\frac{-1}{2}}{x^2+4x+5}\]
\[\frac{3x+4}{(x-3)(x^2+4x+5)}=\frac{1}{2(x-3)}-\frac{x+1}{2(x^2+4x+5)}\]
Hence,
\[\frac{3x+4}{(x-3)(x^2+4x+5)}=\frac{1}{2(x-3)}-\frac{x+1}{2(x^2+4x+5)}.\]
