Question

Integrate with respect to $x$:
$\frac{3x+2}{(x+1{)}^2(x+2)}$

Answer 1

$\int \frac{3x+2}{{(x+1)}^2(x+2)}dxLet,\frac{3x+2}{{(x+1)}^2(x+2)}=\frac{A}{(x+1)}+\frac{B}{{(x+1)}^2}+\frac{C}{(x+2)}\Rightarrow 3x+2=A(x+1)(x+2)+B(x+2)+C{(x+1)}^2\Rightarrow 3x+2=(A+C)x^2+(3A+B+2C)x+(2A+2B+C)$

comparing both side we get;

$(A+C)=0⟶\left(1\right)(3A+B+2C)=3⟶\left(2\right)(2A+2B+C)=2⟶\left(3\right)$

solving the above equation we get the values;

$A=4;B=-1;C=-4$

$now,\int \frac{3x+2}{{(x+1)}^2(x+2)}dx=\int [\frac{4}{(x+1)}-\frac{1}{{(x+1)}^2}-\frac{4}{(x+2)}]dx=4\log (x+1)+\frac{1}{(x+1)}-4\log (x+2)+C=\log {(x+1)}^4+\frac{1}{(x+1)}-\log {(x+2)}^4+C$