burgerandcheese
Junior Member
- Joined
- Jul 2, 2018
- Messages
- 85
The answer is ln(|3x - 1|) /3 + c, but if we know ∫ 1/x dx = ln(|x|) + c, why don't we straight away apply the formula to get ∫ 1/(3x - 1)dx = ln(|3x - 1|) + c ?
If you want to see my working, my initial approach to this problem was by applying the rule "∫ g(ax + b)dx = f(ax + b) /a + c f(x) is the simplest integral of g(x)" and
So g(x) = 1/x, then f(x) = ∫ g(x)dx = ln (|x|)
∫ 1/(3x - 1)dx = ln(|3x - 1|) /3 + c
If you want to see my working, my initial approach to this problem was by applying the rule "∫ g(ax + b)dx = f(ax + b) /a + c f(x) is the simplest integral of g(x)" and
So g(x) = 1/x, then f(x) = ∫ g(x)dx = ln (|x|)
∫ 1/(3x - 1)dx = ln(|3x - 1|) /3 + c