Answer
$\frac{40}{9}+9\ln 3$
Work Step by Step
$\int_1^3\frac{(3x+1)^2}{x^3}dx=\int_1^3\frac{9x^2+6x+1}{x^3}dx$
$=\int_1^3\frac{9}{x}+\frac{6}{x^2}+\frac{1}{x^3}dx$ (Use the sum property for integrals)
$=\int_1^3\frac{9}{x}dx+\int_1^36x^{-2}dx+\int_1^3x^{-3}dx$
$=[9\ln x]_1^3+[\frac{6x^{-2+1}}{-2+1}]_1^3+[\frac{x^{-3+1}}{-3+1}]_1^3$
$=[9\ln x]_1^3+[-\frac{6}{x}]_1^3+[-\frac{1}{2x^2}]_1^3$
$=(9\ln 3-9\ln 1)+(-\frac{6}{3}-(-\frac{6}{1}))+(-\frac{1}{18}-(-\frac{1}{2}))$
$=(9\ln 3- 9\cdot 0)+(-2+6)+(-\frac{1}{18}+\frac{1}{2})$
$=9\ln 3+4+\frac{4}{9}$
$=\frac{40}{9}+9\ln 3$
