$\int \frac{\mathrm{x}+2}{\sqrt{{\mathrm{x}}^2+2\mathrm{x}+3}}\mathrm{dx}$ is equal to 

Let 
$$\begin{gathered} \mathrm{x}+2=\mathrm{A}\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{x}}^2+2\mathrm{x}+3\right)+\mathrm{B} \\ \begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{x}+2=\mathrm{A}(2\mathrm{x}+2)+\mathrm{B}\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em} \mathrm{x}+2=2\mathrm{Ax}+(2\mathrm{A}+\mathrm{B})\end{array} \end{gathered}$$
On equating the coefficient of x and constant term on both sides, we get
      $2\mathrm{A}=1\Rightarrow \mathrm{A}=\frac{1}{2}$
and $2\mathrm{A}+\mathrm{B}=2\Rightarrow 2\times \frac{1}{2}+\mathrm{B}=2$
$$\begin{gathered} \Rightarrow \mathrm{B}=2-1=1 \\ \Rightarrow \mathrm{x}+2=\frac{1}{2}\left(2\mathrm{x}+2\right)+1 \end{gathered}$$
$\begin{array}{l}\therefore \int \frac{\mathrm{x}+2}{\sqrt{{\mathrm{x}}^2+2\mathrm{x}+3}}\mathrm{dx}=\int \frac{\frac{1}{2}(2\mathrm{x}+2)+1}{\sqrt{{\mathrm{x}}^2+2\mathrm{x}+3}}\mathrm{dx}\\ =\frac{1}{2}\int \frac{2\mathrm{x}+2}{\sqrt{{\mathrm{x}}^2+2\mathrm{x}+3}}\mathrm{dx}+\int \frac{\mathrm{dx}}{\sqrt{{\mathrm{x}}^2+2\mathrm{x}+3}}\end{array}$
Let  ${\mathrm{l}}_1=\int \frac{2\mathrm{x}+2}{\sqrt{{\mathrm{x}}^2+2\mathrm{x}+3}}\mathrm{dx}\text{ and }{\mathrm{I}}_2=\int \frac{\mathrm{dx}}{\sqrt{{\mathrm{x}}^2+2\mathrm{x}+3}}$
Then,$\int \frac{\mathrm{x}+2}{\sqrt{{\mathrm{x}}^2+2\mathrm{x}+3}}\mathrm{dx}=\frac{1}{2}{\mathrm{I}}_1+{\mathrm{I}}_2 .....\left(\mathrm{i}\right)$
Now,   ${\mathrm{I}}_1=\int \frac{2\mathrm{x}+2}{\sqrt{{\mathrm{x}}^2+2\mathrm{x}+3}}\mathrm{dx}$
Let ${\mathrm{x}}^2+2\mathrm{x}+3=\mathrm{t}\Rightarrow (2\mathrm{x}+2)\mathrm{dx}=\mathrm{dt}$
      $\begin{array}{l}{\mathrm{I}}_2=\int \frac{\mathrm{dt}}{\sqrt{\mathrm{t}}}=\int {\mathrm{t}}^{-1/2}\mathrm{dt}=\frac{{\mathrm{t}}^{(-1/2)}+1}{1/2+1}+\mathrm{C}=2\sqrt{\mathrm{t}}+{\mathrm{C}}_1\\ =2\sqrt{{\mathrm{x}}^2+2\mathrm{x}+3}+{\mathrm{C}}_1\end{array}$
and
 $\begin{array}{l}{\mathrm{I}}_2=\int \frac{\mathrm{dx}}{\sqrt{{\mathrm{x}}^2+2\mathrm{x}+3}}=\int \frac{\mathrm{dx}}{\sqrt{{\mathrm{x}}^2+2\mathrm{x}+(1{)}^2+3-(1{)}^2}}\\ =\int \frac{\mathrm{dx}}{\sqrt{(\mathrm{x}+1{)}^2+(\sqrt{2}{)}^2}}\end{array}$
Let $\begin{array}{l}\mathrm{x}+1=\mathrm{t}\Rightarrow \mathrm{dx}=\mathrm{dt}\\ {\mathrm{I}}_2=\int \frac{\mathrm{dt}}{\sqrt{{\mathrm{t}}^2+(\sqrt{2}{)}^2}}=\log \left|\mathrm{t}+\sqrt{{\mathrm{t}}^2+2}\right|+{\mathrm{C}}_2\end{array}$
                          $\left[\because \int \frac{\mathrm{dx}}{\sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2}}=\log \mid \mathrm{x}+\sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2}\right]$
$\begin{array}{r}=\log \left|\mathrm{x}+1+\sqrt{(\mathrm{x}+1{)}^2+2}\right|+{\mathrm{C}}_2\\ =\log \left|\mathrm{x}+1+\sqrt{{\mathrm{x}}^2+2\mathrm{x}+3}\right|+{\mathrm{C}}_2\end{array}$
On putting the values of l₁ and l₂ in Eq. (i), we get
$\begin{array}{r}\int \frac{\mathrm{x}+2}{\sqrt{{\mathrm{x}}^2+2\mathrm{x}+3}}\mathrm{dx}=\frac{1}{2}\left[2\sqrt{{\mathrm{x}}^2+2\mathrm{x}+3}\right]\\ +\log \left|\mathrm{x}+1+\sqrt{{\mathrm{x}}^2+2\mathrm{x}+3}\right|\\ \left[\because \frac{1}{2}{\mathrm{C}}_1+{\mathrm{C}}_2=\mathrm{C}\right]\\ =\sqrt{{\mathrm{x}}^2+2\mathrm{x}+3}+\log \mid \mathrm{x}+1+\sqrt{{\mathrm{x}}^2+2\mathrm{x}+3}+\mathrm{C}\end{array}$