Answer
$\frac{1}{3}\ln {4.5}\approx.5014$
Work Step by Step
$\int\limits_2^3\frac{x^2+2x-2}{x^3+3x^2-6x}dx$
Now use the substitution rule for definite integrals
$u=x^3+3x^2-6x$
$dx=\frac{du}{3x^2+6x-6}$
lower limit 2 goes to $(2)^3+3(2)^2-6(2)=8$
upper limit 3 goes to $(3)^3+3(3)^2-6(3)=36$
$\int\limits_8^{36}\frac{x^2+2x-2}{u}\frac{du}{3x^2+6x-6}=\int\limits_8^{36}\frac{1}{3}\frac{1}{u}du=\frac{1}{3}\int\limits_8^{36}\frac{1}{u}du$
$=\frac{1}{3}\ln u|_8^{36}$ (Fundamental theorem)
$=\frac{1}{3}(\ln {36}-\ln 8)=\frac{1}{3}\ln {\frac{36}{8}}=\frac{1}{3}\ln {4.5}\approx.5014$
