Answer
Concave upward on
$$
(-\infty, -3 ), \quad (1, \infty) ,
$$
concave downward on
$$
(-3 , -1) , \quad (-1 , 1)
$$
inflection points at
$$
(-3,9+8 \ln 2 ), \quad (1, 2+8 \ln 2 ).
$$
Work Step by Step
$$
f(x)=x^{2}+8 \ln |x+1|
$$
The first derivative is
$$
\begin{aligned}
f^{\prime}(x) &=(2)x+\frac{8(1)}{(x+1) }\\
&=2x+\frac{8}{(x+1) }
\end{aligned}
$$
and the second derivative is
$$
\begin{aligned}
f^{\prime\prime}(x) &=2-\frac{8(1)}{(x+1)^{2}} \\
&=2-\frac{8}{(x+1)^{2}} \\
\end{aligned}
$$
A function $f$ is concave upward on an interval if $f^{\prime\prime}(x) \gt 0$ for all $x$ in the interval, so
$$
\begin{aligned}
f^{\prime\prime}(x) &=2-\frac{8}{(x+1)^{2}} \gt 0 \\
\Rightarrow\quad\quad\quad\quad\quad\quad\quad\quad\quad\\
-\frac{8}{(x+1)^{2}} & \gt -2\\
8 & \lt 2(x+1)^{2} \\
4 & \lt (x+1)^{2} \\
\Rightarrow\quad\quad\quad\quad\quad\quad\quad\quad\quad\\
x+1 \lt -2 \quad\quad \text {and} & \quad\quad x+1 \gt 2\\
\Rightarrow \quad\quad\quad\quad\quad\quad\quad\quad\quad\\
x \lt -3 \quad\quad \text {and} & \quad\quad x \gt 1
\end{aligned}
$$
so $f(x)$ is concave upward on $(-\infty, -3 ), \quad (1, \infty)$.
Also function $f$ is concave downward on an interval if $f^{\prime\prime}(x) \lt 0$ for all $x$ in the interval, so
$$
\begin{aligned}
f^{\prime\prime}(x) &=2-\frac{8}{(x+1)^{2}} \lt 0 \\
\Rightarrow \quad\quad\quad\quad\quad\quad\quad\quad\quad\\
-\frac{8}{(x+1)^{2}} & \lt -2\\
8 & \gt 2(x+1)^{2} \\
4 & \gt (x+1)^{2} \\
\Rightarrow \quad\quad\quad\quad\quad\quad\quad\quad\quad\\
-2 \lt x+1 & \lt 2\\
\Rightarrow \quad\quad\quad\quad\quad\quad\quad\quad\quad\\
-3 \lt x & \lt 1\\
\end{aligned}
$$
But $f(x)$ is not defined at $x=-1$,
so $f(x)$ is concave downward on $(-3 , -1) $ and $(-1 , 1) $ .
Since the sign of $f^{\prime\prime}(x)$ changes from positive to negative at $ -3 $, the graph changes from concave upward to concave downward at that point, so $(-3 , f(-3)$ is inflection point.
Also, the sign of $f^{\prime\prime}(x)$ changes from negative to positive at $1 $, the graph changes from concave downward to concave upward at that point, so $(1 , f(1)$ is inflection point.
Now ,we find $f( -3), f(1)$as follows:
$$
f(-3)=-3^{2}+8 \ln |-3+1|=9+8 \ln |-2|=9+8 \ln 2,\\
f(1)=(1)^{2}+8 \ln |1+1|=2+8 \ln |2|=2+8 \ln 2,
$$
Finally, we have inflection points at
$$
(-3,9+8 \ln 2 ), \quad (1, 2+8 \ln 2 ).
$$
