# Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.2 Exercises: 13

$\frac{1}{3}$$\ln|x^{3}+3x^{2}+9x| + C #### Work Step by Step \int\frac{x^{2}+2x+3}{x^{3}+3x^{2}+9x} dx Let u =x^{3}+3x^{2}+9x \frac{du}{dx} = 3x^{2}+6x+9 \frac{du}{3x^{2}+6x+9} = dx Substitute u and dx into the original equation \int\frac{x^{2}+2x+3}{u}\frac{du}{3x^{2}+6x+9} = \int\frac{x^{2}+2x+3}{3x^{2}+6x+9}\frac{1}{u} du = \int\frac{x^{2}+2x+3}{3(x^{2}+2x+3)}\frac{1}{u} du = \int\frac{1}{3}\frac{1}{u} du = \frac{1}{3}$$\int\frac{1}{u} du$ = $\frac{1}{3}$$\ln|u| + C Since u =x^{3}+3x^{2}+9x, substituting it back will give you \frac{1}{3}$$\ln|x^{3}+3x^{2}+9x|$ + C

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.