$\int \frac{x+1}{x^{2}+5 x+6} d x=$

Question

Aicharya · Accepted Answer

$I=\int \frac{x+1}{x^{2}+5 x+6} d x$ Let $\quad \frac{x+1}{(x+3)(x+2)}=\int\left[\frac{x+1}{(x+3)(x+2)}ight] d x$ $	herefore x+1=A(x+2)+B(x+3)$ When $x=-2$, we get $B=-1$ When $x=-3$, we get $A=2$ $	herefore I=\int\left[\frac{2}{x+3}-\frac{B}{x+2}ight] d x$ $	herefore 2 \log |(x+3)|-\log |x+2|+c$

$\int \frac{x+1}{x^{2}+5 x+6} d x=$

Solution