∫ 1/(3x - 1)dx

[burgerandcheese]

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

 

[Dr.Peterson]

Have you learned how to do integration by substitution?

Your "rule" is just a special case of that, using the substitution u = ax + b; it can be verified by differentiating the RHS using the chain rule.

Your wrong answer comes from naive "substitution", disregarding how substitution actually works.

 

[burgerandcheese]

Have you learned how to do integration by substitution?

Your "rule" is just a special case of that, using the substitution u = ax + b; it can be verified by differentiating the RHS using the chain rule.

Your wrong answer comes from naive "substitution", disregarding how substitution actually works.

No I don't think I've learned integration by substitution

I think I get it. I don't know how to explain what's in my mind but the x in 1/x is not the same as 3x - 1 because x is a variable. Is my logic correct?

 

[pka]

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 ?

No I don't think I've learned integration by substitution. I don't know how to explain what's in my mind but the x in 1/x is not the same as 3x - 1 because x is a variable. Is my logic correct?

Here is my answer to your first post.
Consider these two functions: i) \(\displaystyle g(x)=\ln(|3x-1|)\) ii) \(\displaystyle h(x)=\frac{\ln(|3x-1|)}{3}\).
Please find each of these derivatives: i) \(\displaystyle g'(x)=~?\) i) \(\displaystyle h'(x)=~?\)

You may well wonder about the derivative of \(\displaystyle \ln(|x|)\).
\(\displaystyle \ln(|x|)=\begin{cases}\ln(x) &: x>0 \\ \ln(-x) &: x<0\end{cases}\;\;\text{Please note }x\ne 0.\)____________\(\displaystyle \frac{d}{dx}\left\{\ln(|x|)\right\}=\)\(\displaystyle \begin{cases}\dfrac{1}{x} &: x>0 \\ \dfrac{-1}{-x}=\dfrac{1}{x} &: x<0\end{cases} \)

 

[Steven G]

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

You let g(x) = (3x-1). THEN g'(x) = 3 or dx = g'(x)dx/3. That is where you lost the /3.
Most students are taught to write u instead of g(x), so u = 3x-1 and du = 3dx or dx = du/3. Substutite this into your original integral and then factor out the 1/3. Then all will work out.

 

[burgerandcheese]

Here is my answer to your first post.
Consider these two functions: i) \(\displaystyle g(x)=\ln(|3x-1|)\) ii) \(\displaystyle h(x)=\frac{\ln(|3x-1|)}{3}\).
Please find each of these derivatives: i) \(\displaystyle g'(x)=~?\) i) \(\displaystyle h'(x)=~?\)

You may well wonder about the derivative of \(\displaystyle \ln(|x|)\).
\(\displaystyle \ln(|x|)=\begin{cases}\ln(x) &: x>0 \\ \ln(-x) &: x<0\end{cases}\;\;\text{Please note }x\ne 0.\)____________\(\displaystyle \frac{d}{dx}\left\{\ln(|x|)\right\}=\)\(\displaystyle \begin{cases}\dfrac{1}{x} &: x>0 \\ \dfrac{-1}{-x}=\dfrac{1}{x} &: x<0\end{cases} \)

Thank you. I got g'(x) = 3/(3x - 1) and h'(x) = 1/(3x - 1)

I wanted to use d/dx ln(|x|) = 1/x but I didn't know how to so I split it into two cases for x < 0 and x > 0. But now I know I had to let u = 3x - 1 and not u = |3x - 1|

 

[pka]

Thank you. I got g'(x) = 3/(3x - 1) and h'(x) = 1/(3x - 1)

CORRECT! Good for you, way to go.

Now which of those answers \(\displaystyle \int {\frac{{dx}}{{3x - 1}} = ?} \)

 

[burgerandcheese]

CORRECT! Good for you, way to go.

Now which of those answers \(\displaystyle \int {\frac{{dx}}{{3x - 1}} = ?} \)

certainly ln(|3x - 1|)/3 ?

You let g(x) = (3x-1). THEN g'(x) = 3 or dx = g'(x)dx/3. That is where you lost the /3.
Most students are taught to write u instead of g(x), so u = 3x-1 and du = 3dx or dx = du/3. Substutite this into your original integral and then factor out the 1/3. Then all will work out.

Wow! I don't think I've learnt this yet