Question

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

Answer 1

$\int \frac{1}{(x+1)(x+2)(x+3)}dxlet,\frac{1}{(x+1)(x+2)(x+3)}=\frac{A}{x+1}+\frac{B}{x+2}+\frac{C}{x+3}\Rightarrow 1=(A+B+C)x^2+(5A+4B+3C)x+(6A+3B+2C)$

comparing both side we get;

$(A+B+C)=0⟶\left(1\right)(5A+4B+3C)=0⟶\left(2\right)(6A+3B+2C)=1⟶\left(3\right)$

solving the above equation we get the value;

$A=-\frac{1}{2};B=1;C=\frac{1}{2}now,\int \frac{1}{(x+1)(x+2)(x+3)}dx=\int [\frac{-1}{2(x+1)}+\frac{1}{x+2}+\frac{1}{2(x+3)}]dx=\frac{-1}{2}\log |x+1|+\log |x+2|+\frac{1}{2}\log |x+3|+C=\log {|x+1|}^{-\frac{1}{2}}+\log |x+2|+\log {|x+3|}^{\frac{1}{2}}+C=\log \left|\frac{(x+2){(x+3)}^{\frac{1}{2}}}{{(x+1)}^{\frac{1}{2}}}\right|+C$