Let $\text{I}=\int\frac{(3\text{x}+5)}{\text{x}^2+3\text{x}-18}$
$\text{I}=\int\frac{(3\text{x}+5)\text{dx}}{(\text{x}+6)(\text{x}-3)}$
Let $\int\frac{(3\text{x}+5)}{(\text{x}+6)(\text{x}-3)}=\frac{\text{A}}{\text{x}+6}+\frac{\text{B}}{\text{x}-3}$
So, $3\text{x}+5=\text{A}(\text{x}-3)+\text{B}(\text{x}+6)$
On comparing,
$\text{A}+\text{B}=3\ \dots(\text{i})$
$-3\text{A}+6\text{B}=5\ \dots(\text{ii})$
$-3\text{A}+6(3-\text{A})=5$
$-3\text{A}+18-6\text{A}=5$
$\text{A}=\frac{-13}{9}=\frac{13}{9}$ and $\text{B}=3-\text{A}=3-\frac{13}{9}=\frac{14}{9}$
So, $\int\frac{(3\text{x}+5)\text{dx}}{(\text{x}+6)(\text{x}-3)}$
$=\int\frac{13\text{dx}}{9(\text{x}+6)}+\int\frac{14\text{dx}}{9(\text{x}-3)}$
$=\frac{13}{9}\text{ln}(\text{x}+6)+\frac{14}{9}\text{ln}(\text{x}-3)+\text{C}$
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