Evaluate the integral:

$I=\int \frac{x+1}{x^{2}+x+3} d x$

As we can see that there is a term of $x$ in numerator and derivative of $x^{2}$ is also $2 x$. So there is a chance that we can make substitution for $x^{2}+x+3$ and I can be reduced to a fundamental integration.

As, $\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{2}+\mathrm{x}+1\right)=2 \mathrm{x}+1$

$\therefore$ Let, $x=A(2 x+1)+B$

$\Rightarrow x=2 A x+A+B$

On comparing both sides -

We have,

$2 \mathrm{~A}=1 \Rightarrow \mathrm{A}=1 / 2$

$A+B=0 \Rightarrow B=-A=-1 / 2$

Hence,

$I=\int \frac{\frac{1}{2}(2 x+1)-\frac{1}{2}}{x^{2}+x+3} d x$

$\therefore I=\frac{1}{2} \int \frac{2 x+1}{x^{2}+x+3} d x-\frac{1}{2} \int \frac{1}{x^{2}+x+3} d x$

Let, $I_{1}=\frac{1}{2} \int \frac{2 x+1}{x^{2}+x+3} d x$ and $I_{2}=\frac{1}{2} \int \frac{1}{x^{2}+x+3} d x$

Now, $I=I_{1}-I_{2} \ldots .$ eqn 1

We will solve $I_{1}$ and $I_{2}$ individually.

As $I_{1}=\frac{1}{2} \int \frac{2 x+1}{x^{2}+x+3} d x$

Let $u=x^{2}+x+3 \Rightarrow d u=(2 x+1) d x$

$\therefore \mathrm{I}_{1}$ reduces to $\frac{1}{2} \int \frac{\mathrm{du}}{\mathrm{u}}$

Hence,

$\mathrm{I}_{1}=\frac{1}{2} \int \frac{\mathrm{du}}{\mathrm{u}}=\frac{1}{2} \log |\mathrm{u}|+\mathrm{C}\left\{\because \int \frac{\mathrm{dx}}{\mathrm{x}}=\log |\mathrm{x}|+\mathrm{C}\right\}$

On substituting the value of $u$, we have:

$I_{1}=\frac{1}{2} \log \left|x^{2}+x+3\right|+C \ldots . .$ eqn 2

As, $I_{2}=\frac{1}{2} \int \frac{1}{x^{2}+x+3} d x$ and we don't have any derivative of function present in denominator. $\therefore$ we will use some special integrals to solve the problem.

As denominator doesn't have any square root term. So one of the following two integrals will solve the problem.

i) $\int \frac{1}{x^{2}-a^{2}} d x=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+C$ ii) $\int \frac{1}{x^{2}+a^{2}} d x=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+C$

Now we have to reduce $I_{2}$ such that it matches with any of above two forms.

We will make to create a complete square so that no individual term of $x$ is seen in denominator.

$\therefore I_{2}=\frac{1}{2} \int \frac{1}{\mathrm{x}^{2}+\mathrm{x}+3} \mathrm{dx}$

$\Rightarrow I_{2}=\frac{1}{2} \int \frac{1}{\left\{x^{2}+2\left(\frac{1}{2}\right) x+\left(\frac{1}{2}\right)^{2}\right\}+3-\left(\frac{1}{2}\right)^{2}} d x$

Using: $a^{2}+2 a b+b^{2}=(a+b)^{2}$

We have:

$I_{2}=\frac{1}{2} \int \frac{1}{\left(x+\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{11}}{2}\right)^{2}} d x$

$\mathrm{I}_{2}$ matches with $\int \frac{1}{\mathrm{x}^{2}+\mathrm{a}^{2}} \mathrm{dx}=\frac{1}{\mathrm{a}} \tan ^{-1}\left(\frac{\mathrm{x}}{\mathrm{a}}\right)+\mathrm{C}$

$\therefore I_{2}=\frac{1}{2}\left\{\frac{1}{\left(\frac{\sqrt{11}}{2}\right)} \tan ^{-1}\left(\frac{x+\frac{1}{2}}{\frac{\sqrt{11}}{2}}\right)+C\right\}$

$\Rightarrow I_{2}=\frac{1}{\sqrt{11}} \tan ^{-1}\left(\frac{2 x+1}{\sqrt{11}}\right)+C \ldots$ eqn 3

From eqn 1:

$I=I_{1}-I_{2}$

Using eqn 2 and eqn 3:

$I=\frac{1}{2} \log \left|x^{2}+x+3\right|+\frac{1}{\sqrt{11}} \tan ^{-1}\left(\frac{2 x+1}{\sqrt{11}}\right)+C$