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Answer:

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

Hint: Using Formula $\int \frac{d x}{x^{2}-a^{2}}=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+c \& \int \frac{d t}{t}=\log t+c$

Given:  $\int \frac{2 x-3}{x^{2}+6 x+13} d x$

Solution:          $\frac{d}{d x}\left(x^{2}+6 x+13\right)=2 x+6$

Let,           $2 x-3=A(2 x+6)+B$

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

On comparing,

$2 x=2 A x=>A=1$

$6 A+B=-3=>B=-9$

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

$\text { Let, } x^{2}+6 x+13=t$

$(2 x+6) d x=d t$

$I=\int \frac{d t}{t}-9 \int \frac{d x}{x^{2}+2(3) x+3^{2}-3^{2}+13}$

$I=\int \frac{d t}{t}-9 \int \frac{d x}{(x+3)^{2}-9+13}$

$I=\int \frac{d t}{t}-9 \int \frac{d x}{(x+3)^{2}-4}$

$I=\int \frac{d t}{t}-9 \int \frac{d x}{(x+3)^{2}-2^{2}}$                                                    $\left[\int \frac{d x}{x^{2}-a^{2}}=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+c \& \int \frac{d t}{t}=\log t+c\right]$

$I=\log |t|-9 \times \frac{1}{2} \tan ^{-1}\left(\frac{x+3}{2}\right)+c$

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

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