Split up the integral:
[tex]\displaystyle\int\frac{x+2019}{x^2+9}\,\mathrm dx = \int\frac{x}{x^2+9}\,\mathrm dx + \int\frac{2019}{x^2+9}\,\mathrm dx[/tex]
For the first integral, substitute y = x ² + 9 and dy = 2x dx. For the second integral, take x = 3 tan(z) and dx = 3 sec²(z) dz. Then you get
[tex]\displaystyle \int\frac x{x^2+9}\,\mathrm dx = \frac12\int{2x}{x^2+9}\,\mathrm dx \\\\ = \frac12\int\frac{\mathrm du}u \\\\ = \frac12\ln|u| + C \\\\ =\frac12\ln\left(x^2+9\right)[/tex]
and
[tex]\displaystyle \int\frac{2019}{x^2+9}\,\mathrm dx = 2019\int\frac{3\sec^2(z)}{(3\tan(z))^2+9}\,\mathrm dz \\\\ = 2019\int\frac{3\sec^2(z)}{9\tan^2(z)+9}\,\mathrm dz \\\\ = 673\int\frac{\sec^2(z)}{\tan^2(z)+1}\,\mathrm dz \\\\ = 673\int\frac{\sec^2(z)}{\sec^2(z)}\,\mathrm dz \\\\ = 673\int\mathrm dz \\\\ = 673z+C \\\\ = 673\arctan\left(\frac x3\right)+C[/tex]
Then
[tex]\displaystyle\int\frac{x+2019}{x^2+9}\,\mathrm dx = \boxed{\frac12\ln\left(x^2+9\right) + 673\arctan\left(\frac x3\right) + C}[/tex]
