how to partial fraction $\frac{1}{(x+1)^2}$ I need to integrate $\frac{1}{(x^2+2x+1)}$, so I need to use partial fraction as the polynomial can be factored as $\frac{1}{(x+1)^2}$. This is what I've tried:
$$\frac{A}{(x+1)} + \frac{B}{(x+1)^2}$$
$$A\cdot(x+1)^2 + B\cdot(x+1)$$
$$Ax^2+2Ax+A+Bx+B$$
$$Ax^2+(2A+B)x+(A+B)$$
So,
$$A=0$$
$$2A+B=0$$
$$A+B=1$$
but that does not make sense, because if $A=0$, then $2A+B$ can't be zero, could you please tell me what's the problem?
 A: The fraction becomes
$$
\frac{A}{(x+1)} + \frac{B}{(x+1)^2}=
\frac{A(x+1)+B}{(x+1)^2}=
\frac{Ax+(A+B)}{(x+1)^2}
$$
so $A=0$ and $B=1$. Indeed
$$
\frac{1}{(x+1)^2}
$$
is already in “partial fractions” form. And
$$
\int\frac{1}{(x+1)^2}\,dx=-\frac{1}{x+1}+C
$$

It would be different if you started with
$$
\frac{x}{(x+1)^2}
$$
because then the decomposition would give $A=1$ and $B=-1$, so
$$
\int\frac{x}{(x+1)^2}\,dx=
\int\left(\frac{1}{x+1}-\frac{1}{(x+1)^2}\right)\,dx=
\log|x+1|+\frac{1}{x+1}+C
$$

Where's the problem with your computations? You have done
$$
\frac{A}{(x+1)} + \frac{B}{(x+1)^2}=
\frac{A(x+1)^2+B(x+1)}{(x+1)^3}
$$
and then equalled this with $\frac{1}{(x+1)^2}$ without taking into account the different denominators.
A: Just lookoing at the line $\frac{A}{(x+1)} + \frac{B}{(x+1)^2} = \frac{1}{(x+1)^2}$ we clearly see that $A = 0, B = 1$. As for your mistake, you multiplied by $(x+1)$ one time too many when removing the denominators.
A: You can integred it directly without partial fraction as $$\int { \frac { dx }{ { \left( x+1 \right)  }^{ 2 } } =\int { \frac { d\left( x+1 \right)  }{ { \left( x+1 \right)  }^{ 2 } }  } =-\frac { 1 }{ x+1 } +C } $$
