Evaluate $\int \frac{\cos \mathrm{x}+3\sin \mathrm{x}+7}{\cos \mathrm{x}+\sin \mathrm{x}+1}\mathrm{dx}$

Let $\mathrm{I}=\int \frac{\cos \mathrm{x}+3\sin \mathrm{x}+7}{\cos \mathrm{x}+\sin \mathrm{x}+1}\mathrm{dx}$
Let $\cos \mathrm{x}+3\sin \mathrm{x}+7\equiv \mathrm{A}(\cos \mathrm{x}+\sin \mathrm{x}+1) +\mathrm{B}\frac{\mathrm{d}}{\mathrm{dx}}(\cos \mathrm{x}+\sin \mathrm{x}+1)+\mathrm{C}$

comparing coefficients of cosx,sinx and constant terms on both sides
$\begin{array}{l}\Rightarrow \mathrm{A}+\mathrm{B}=1,\mathrm{A}-\mathrm{B}=3\text{ and }\mathrm{A}+\mathrm{C}=7\\ \Rightarrow \mathrm{A}=2,\mathrm{B}=-1,\mathrm{C}=5\\ \therefore \mathrm{I}=2\int 1\mathrm{dx}+(-1)\int \frac{(-\sin \mathrm{x}+\cos \mathrm{x})}{\cos \mathrm{x}+\sin \mathrm{x}+1}\mathrm{dx}\\ +5\int \frac{1}{\cos \mathrm{x}+\sin \mathrm{x}+1}\mathrm{dx}\end{array}$
$\text{ Put }\tan \left(\frac{\mathrm{x}}{2}\right)=\mathrm{t}\Rightarrow \mathrm{dx}=\frac{2\mathrm{dt}}{1+{\mathrm{t}}^2}$
$\begin{array}{l}\sin \mathrm{x}=\frac{2\mathrm{t}}{1+{\mathrm{t}}^2},\cos \mathrm{x}=\frac{1-{\mathrm{t}}^2}{1+{\mathrm{t}}^2}\\ \int \frac{1}{1+\sin \mathrm{x}+\cos \mathrm{x}}\mathrm{dx}\\ =\int \frac{2\mathrm{dt}}{\left(1+{\mathrm{t}}^2\right)+2\mathrm{t}+\left(1-{\mathrm{t}}^2\right)}=\int \frac{\mathrm{dt}}{1+\mathrm{t}}\\ =\log |1+\mathrm{t}|+\mathrm{c}=\log \left|1+\tan \frac{\mathrm{x}}{2}\right|+\mathrm{c}\\ \therefore \mathrm{I}=2\mathrm{x}-\log |\cos \mathrm{x}+\sin \mathrm{x}+1|+5\log \left|1+\tan \frac{\mathrm{x}}{2}\right|+\mathrm{k}\end{array}$
Where k is a integration constant