Evaluate the integral:

$I=\int \frac{2 x-3}{x^{2}+6 x+13} d x$

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 a substitution for $x^{2}+6 x+13$ and I can be reduced to a fundamental integration.

As $\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{2}+6 \mathrm{x}+13\right)=2 \mathrm{x}+6$

$\therefore$ Let, $2 x-3=A(2 x+6)+B$

$\Rightarrow 2 x-3=2 A x+6 A+B$

On comparing both sides -

We have,

$2 A=2 \Rightarrow A=1$

$6 A+B=-3 \Rightarrow B=-3-6 A=-9$

Hence,

$I=\int \frac{(2 x+6)-9}{x^{2}+6 x+13} d x$

$\therefore I=\int \frac{2 x+6}{x^{2}+6 x+13} d x-9 \int \frac{1}{x^{2}+6 x+13} d x$

Let, $I_{1}=\int \frac{2 x+6}{x^{2}+6 x+13} d x$ and $I_{2}=\int \frac{1}{x^{2}+6 x+13} d x$

Now, $I=I_{1}-9 I_{2} \ldots$ eqn $I$

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

As, $I_{1}=\int \frac{2 x+6}{x^{2}+6 x+13} d x$

Let $u=x^{2}+6 x+13 \Rightarrow d u=(2 x+6) d x$

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

Hence,

$I_{1}=\int \frac{d u}{u}=\log |u|+C\left\{\because \int \frac{d x}{x}=\log |x|+C\right\}$

On substituting value of $u$, we have:

$I_{1}=\log \left|x^{2}+6 x+13\right|+C \ldots$ eqn 2

As, $I_{2}=\int \frac{1}{x^{2}+6 x+13} d x$ 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.

$\therefore I_{2}=\int \frac{1}{x^{2}+6 x+13} d x$

$\Rightarrow I_{2}=\int \frac{1}{\left\{x^{2}+2(3) x+(3)^{2}\right\}+13-(3)^{2}} d x$

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

We have:

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

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

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

From eqn 1:

$I=I_{1}-9 I_{2}$

Using eqn 2 and eqn 3 :

$I=\log \left|x^{2}+6 x+13\right|-9 \frac{1}{2} \tan ^{-1}\left(\frac{x+3}{2}\right)+C$

$I=\log \left|x^{2}+6 x+13\right|-\frac{9}{2} \tan ^{-1}\left(\frac{x+3}{2}\right)+C$