Answer::
[tex]\frac{\ln |x|}{2}+\frac{\ln |x^{}+6|}{2}+C[/tex]
Explanation:
Given:
[tex]I=\int \frac{x+3}{x^{2}+6 x}dx[/tex][tex]\begin{gathered} \text{Let }u=x^2+6x \\ \implies\frac{du}{dx}=2x+6=2(x+3) \\ \implies dx=\frac{du}{2(x+3)} \end{gathered}[/tex]
Substitute u and dx into I.
[tex]\begin{gathered} I=\int \frac{x+3}{x^2+6x}dx \\ \implies I=\int \frac{x+3}{u}\frac{du}{2(x+3)}=\int \frac{1}{u}\frac{du}{2}=\frac{1}{2}\int \frac{1}{u}du \end{gathered}[/tex]
The integral of 1/u is the standard integral ln |u|.
Therefore:
[tex]I=\frac{1}{2}\ln |u|+C,C\text{ a constant of integration}[/tex]
SUbstitute back the expression for u.
[tex]\begin{gathered} I=\frac{1}{2}\ln |x^2+6x|+C=\frac{\ln|x||x^{}+6|}{2}+C \\ =\frac{\ln|x|}{2}+\frac{\ln |x+6|}{2}+C \end{gathered}[/tex]
The result of the given integral is:
[tex]\frac{\ln |x|}{2}+\frac{\ln |x^{}+6|}{2}+C[/tex]
