Question

∫
`(x+2019)/(x^(2)+9 )`

Answer 1

Split up the integral:

`\displaystyle\int(x+2019)/(x^2+9)\,\mathrm dx = \int(x)/(x^2+9)\,\mathrm dx + \int(2019)/(x^2+9)\,\mathrm dx`

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

`\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)`

and

`\displaystyle \int(2019)/(x^2+9)\,\mathrm dx = 2019\int(3\sec^2(z))/((3\tan(z))^2+9)\,\mathrm dz \\\\ = 2019\int(3\sec^2(z))/(9\tan^2(z)+9)\,\mathrm dz \\\\ = 673\int(\sec^2(z))/(\tan^2(z)+1)\,\mathrm dz \\\\ = 673\int(\sec^2(z))/(\sec^2(z))\,\mathrm dz \\\\ = 673\int\mathrm dz \\\\ = 673z+C \\\\ = 673\arctan\left(\frac x3\right)+C`

Then

`\displaystyle\int(x+2019)/(x^2+9)\,\mathrm dx = \boxed{\frac12\ln\left(x^2+9\right) + 673\arctan\left(\frac x3\right) + C}`