Answer
\[(a) \frac{A}{1+2x}+\frac{B}{3-x}\]
\[(b) \frac{A}{x}+\frac{B}{x^2}+\frac{C}{x^3}+\frac{D}{1+x}\]
Work Step by Step
\[(a) \frac{4+x}{(1+2x)(3-x)}\]
Since factors in denominator are linear so partial fraction decompositition is given by
\[\frac{4+x}{(1+2x)(3-x)}=\frac{A}{1+2x}+\frac{B}{3-x}\]
\[(b)\frac{1-x}{x^3+x^4}\]
\[\Rightarrow \frac{1-x}{x^3+x^4}= \frac{1-x}{x^3(1+x)}\]
So partial fraction decomposition is given by \[ \frac{1-x}{x^3+x^4}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x^3}+\frac{D}{1+x}\]
Hence , \[\frac{1-x}{x^3+x^4}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x^3}+\frac{D}{1+x}\]
