We can solve this integral using u-substitution.
Let u = x^3, then du/dx = 3x^2, or dx = du/3x^2.
Replacing x and dx with u and du/3x^2, respectively, we get:
∫(x-x³)/x³ dx = ∫(1-u)/u du/3x^2
= (1/3) ∫(u^-1 - 1) du
= (1/3) ln|u| - (1/3)u + C
Substituting back x^3 for u, we get:
(1/3) ln|x^3| - (1/3)x^3 + C
Simplifying this expression, we get:
(1/3) ln|x³| - (1/3)x³ + C
Sigma x-x³/x³ dx
3 answers
and of course,
(1/3) ln|x³ = ln|x|
(1/3) ln|x³ = ln|x|
Yes, that's correct! You can simplify (1/3) ln|x³| to ln|x| by using the property of logarithms that states ln(a^b) = b ln(a), so ln|x³| = 3 ln|x|, and (1/3) ln|x³| simplifies to ln|x|.