Question

$\int \frac{x+1}{x^2+x+3}dx$

Answer 1

$$\begin{gathered} \int \frac{\left(x+1\right)dx}{x^2+x+3} \\ x+1=\frac{\mathrm{A}d}{dx}\left(x^2+x+3\right)+\mathrm{B} \\ x+1=\mathrm{A}\left(2x+1\right)+\mathrm{B} \\ x+1=2\mathrm{A}x+\mathrm{A}+\mathrm{B} \end{gathered}$$

Comparing Coefficients of like powers of x
$$\begin{gathered} 2\mathrm{A}=1 \\ \mathrm{A}=\frac{1}{2} \\ \mathrm{A}+\mathrm{B}=1 \\ \frac{1}{2}+\mathrm{B}=1 \\ \mathrm{B}=\frac{1}{2} \\ \left(x+1\right)=\frac{1}{2}\left(2x+1\right)+\frac{1}{2} \end{gathered}$$

$$\begin{gathered} \mathrm{Now},\int \frac{\left(x+1\right)dx}{x^2+x+3} \\ =\int \frac{{\displaystyle \frac{1}{2}}\left(2x+1\right)dx}{x^2+x+3}+\frac{1}{2}\int \frac{dx}{x^2+x+3} \\ =\frac{1}{2}\int \frac{\left(2x+1\right)dx}{x^2+x+3}+\frac{1}{2}\int \frac{dx}{x^2+x+{\left({\displaystyle \frac{1}{2}}\right)}^2-{\left({\displaystyle \frac{1}{2}}\right)}^2+3} \\ =\frac{1}{2}\int \frac{\left(2x+1\right)dx}{x^2+x+3}+\frac{1}{2}\int \frac{dx}{{\left(x+{\displaystyle \frac{1}{2}}\right)}^2+{\displaystyle 3}-{\displaystyle \frac{1}{4}}} \\ =\frac{1}{2}\int \frac{\left(2x+1\right)dx}{x^2+x+3}+\frac{1}{2}\int \frac{dx}{{\left(x+{\displaystyle \frac{1}{2}}\right)}^2+{\left({\displaystyle \frac{\sqrt{11}}{2}}\right)}^2} \\ =\frac{1}{2}\log \left|x^2+x+3\right|+\frac{1}{2}\times \frac{2}{\sqrt{11}}{\tan }^{-1}\left(\frac{x+{\displaystyle \frac{1}{2}}}{{\displaystyle \frac{\sqrt{11}}{2}}}\right)+C \\ =\frac{1}{2}\log \left|x^2+x+3\right|+\frac{1}{\sqrt{11}}{\tan }^{-1}\left(\frac{2x+1}{\sqrt{11}}\right)+C \end{gathered}$$