Evaluate $\displaystyle \int \dfrac{dx}{\left(x + 1\right)^2 \sqrt{x^2 + 2 x + 2}}$
$\begin{aligned}
\text{Let } I & = \int \dfrac{dx}{\left(x + 1\right)^2 \sqrt{x^2 + 2x +2}} \\\\
& = \int \dfrac{dx}{\left(x + 1\right)^2 \sqrt{\left(x^2 + 2x + 1\right) + 1}} \\\\
& = \int \dfrac{dx}{\left(x + 1\right)^2 \sqrt{\left(x + 1\right)^2 + 1}} \;\;\; \cdots (1)
\end{aligned}$
Let $\dfrac{1}{x + 1} = p$ $\;\;\; \cdots$ (2a)
Differentiating equation (2a) gives
$- \dfrac{1}{\left(x + 1\right)^2} \; dx = dp$ $\implies$ $\dfrac{dx}{\left(x + 1\right)^2} = - dp$ $\;\;\; \cdots$ (2b)
From equation (2a) we have $x + 1 = \dfrac{1}{p}$ $\;\;\; \cdots$ (2c)
In view of equations (2a), (2b) and (2c), equation (1) becomes
$\begin{aligned}
I & = \int \dfrac{-dp}{\sqrt{\dfrac{1}{p^2} + 1}} \\\\
& = - \int \dfrac{p \; dp}{\sqrt{p^2 + 1}} \;\;\; \cdots (3)
\end{aligned}$
Let $p^2 + 1 = t$ $\;\;\; \cdots$ (4a)
Differentiating equation (4a) gives
$2 \; p \; dp = dt$ $\implies$ $p \; dp = \dfrac{dt}{2}$ $\;\;\; \cdots$ (4b)
In view of equations (4a) and (4b), equation (3) becomes
$\begin{aligned}
I & = - \dfrac{1}{2} \int \dfrac{dt}{\sqrt{t}} \\\\
& = - \dfrac{1}{2} \times 2 \times t^{1/2} + c \\\\
& = - \sqrt{p^2 + 1} + c \;\;\; \left[\text{From equation (4a)}\right] \\\\
& = - \sqrt{\left(\dfrac{1}{x + 1}\right)^2 + 1} + c \;\;\; \left[\text{From equation (2a)}\right] \\\\
& = \dfrac{- \sqrt{x^2 + 2x + 2}}{x + 1} + c
\end{aligned}$