2022-M2-06

$\begin{array}{rcl}
I & = & \dint \dfrac{1}{x^2+2x+5}dx \\
I & = & \dint \dfrac{1}{x^2+2x+1+4}dx \\
I & = & \dint \dfrac{1}{(x+1)^2+2^2} dx
\end{array}$

Let $x+1=2\tan\theta$. Then $dx=2\sec^2\theta d\theta$.

$\begin{array}{rcl}
I & = & \dint \dfrac{1}{(2\tan\theta)^2+4} \times 2\sec^2\theta d\theta \\
I & = & \dint \dfrac{\sec^2\theta d\theta}{2(\tan^2\theta +1)} \\
I & = & \dint \dfrac{\sec^2\theta d\theta}{2\sec^2\theta} \\
I & = & \dint \dfrac{1}{2} d\theta \\
I & = & \dfrac{1}{2}\theta +C \text{ , where $C$ is a constant.} \\
I & = & \dfrac{1}{2} \tan^{-1}\left(\dfrac{x+1}{2}\right) +C
\end{array}$

$\begin{array}{rcl}
\dfrac{dy}{dx} & = & \dfrac{2x+1}{x^2+2x+5} \\
y & = & \dint \dfrac{2x+1}{x^2+2x+5} dx \\
y & = & \dint \dfrac{2x+2}{x^2+2x+5}dx -\dint \dfrac{1}{x^2+2x+5}dx \\
y & = & \dint \dfrac{d(x^2+2x+5)}{x^2+2x+5} – \dint \dfrac{1}{x^2+2x+5} dx \\
y & = & \ln \left| x^2+2x+5\right| -\dfrac{1}{2}\tan^{-1}\left(\dfrac{x+1}{2}\right) +C \text{ , by (a).}
\end{array}$

Since $G$ passes through $(-3,\ln 2)$, then sub. $x=-3$ and $y=\ln 2$ into the equation of $G$, we have

$\begin{array}{rcl}
\ln 2 & = & \ln \left| (-3)^2+2(-3)+5\right|-\dfrac{1}{2}\tan^{-1}\left( \dfrac{-3+1}{2}\right) +C \\
\ln 2 & = & \ln 8 -\dfrac{1}{2} \times \dfrac{-\pi}{4} +C \\
C & = & \ln 2-\ln 8-\dfrac{\pi}{8} \\
C & = & \ln 2-\ln2^3 -\dfrac{\pi}{8} \\
C & = & \ln 2-3\ln 2-\dfrac{\pi}{8} \\
C & = & -2\ln 2-\dfrac{\pi}{8}
\end{array}$

Therefore, the equation of $G$ is $y=\ln \left|x^2+2x+5\right|-\dfrac{1}{2}\tan^{-1}\left(\dfrac{x+1}{2}\right)-2\ln 2-\dfrac{\pi}{8}$.

Sub. $x=-1$ into the equation of $G$, we have

$\begin{array}{rcl}
y & = & \ln \left|(-1)^2+2(-1)+5 \right|-\dfrac{1}{2}\tan^{-1}\left(\dfrac{-1+1}{2}\right)-2\ln 2-\dfrac{\pi}{8} \\
y & = & \ln 4-2\ln 2-\dfrac{\pi}{8} \\
y & = & \ln2^2-2\ln 2-\dfrac{\pi}{8} \\
y & = & 2\ln 2-2\ln 2 -\dfrac{\pi}{8} \\
y & = & \dfrac{-\pi}{8}
\end{array}$

Therefore, $G$ passes through the point $\left(-1, \dfrac{-\pi}{8} \right)$.