$\int \frac{2x+1}{\left( 1-x \right)\left( 4+x \right)}dx$
Putting $\frac{2x+1}{\left( 1-x \right)\left( 4+x \right)}dx=\frac{A}{1-x}+\frac{B}{4+x}……(1)$
$A\left( 4+x \right)+B(1-x)=2x+1$
Now put, $1-x=0$
Therefore, $x=1$
$A(5)+B(0)=3$
$A=\frac{3}{5}$
Now put $4+x=0$
Therefore, $x=-4$
$A\left( 0 \right)+B\left( 5 \right)=-8+1=7$
$B=\frac{-7}{5}$
Now from equation (1) we get,
$\frac{2x+1}{(1-x)(4+x)}=\frac{3}{5}\times \frac{1}{1-x}+\frac{-7}{5}\times \frac{1}{4+x}$
\[\int \frac{2x+1}{\left( 1-x \right)\left( 4+x \right)}dx=\frac{3}{5}\int \frac{1}{1-x}dx+\frac{-7}{5}\int \frac{1}{4+x}dx\]
\[=\frac{-3}{5}\log \left| 1-x \right|-\frac{7}{5}\log \left| 4+x \right|+c\]
\[=-\frac{1}{5}\left[ 3\log \left| 1-x \right|+7\log \left| 4+x \right| \right]+c\]