Integral Example
- Thread starter Jason76
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What is the derivative of \(\displaystyle \ln(x)\)? Work backwards to know when the integral will be \(\displaystyle \ln|x|\)
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How are you getting that the derivative of ln(3x) is 3/x?
How are you getting that the derivative of ln(3x) is 3/x?
\(\displaystyle \dfrac{d}{dx} \ln 3x \rightarrow \dfrac{3}{x}\)
So
\(\displaystyle \int \dfrac{3}{x} \rightarrow \ln |3x| + C\) :?:
pka
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HallsofIvy
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Notice, by the way, that ln(3x)= ln(x)+ ln(3). Another reason why \(\displaystyle \frac{d(ln(3x))}{dx}= \frac{d(ln(x))}{dx}= \frac{1}{x}\)
D
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I give up!!
\(\displaystyle \displaystyle \int \frac{3}{x} dx \ = \ 3* \int \frac{1}{x} dx \ = 3 * ln(|x|) + C = ln(|x^3|) + C\)
\(\displaystyle \displaystyle \int \frac{3}{x} dx \ = \ 3* \int \frac{1}{x} dx \ = 3 * ln(|x|) + C = ln(|x^3|) + C\)
A feeling a lot of us have regarding, well, you know.
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Repeating the fact that you'd somehow arrived at a wrong result does not explain how you arrived at that wrong result. Please show your work, especially when it is asked that you do. Thank you!
HallsofIvy
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Surely you learned, long ago, that \(\displaystyle \int Af(x)dx= A\int f(x)dx\) for any constant, A?
\(\displaystyle -\int\dfrac{3}{x}dx= -3\int\dfrac{1}{x}dx\).
Sorry if i made anyone mad on here. I see what the answer is. I will do some work and see if I can understand the mechanics behind it. Also, sorry i hadn't replied sooner, but I was busy doing homework all day.
\(\displaystyle \int \dfrac{1}{x} dx \rightarrow \ln|u| + C\)
So considering \(\displaystyle \int \dfrac{3}{x} dx\) then what is\(\displaystyle u\)?
\(\displaystyle u\) must be \(\displaystyle x^{3}\) but how did it get there?
Perhaps it can be understood as
\(\displaystyle \int 3x^{-1} \rightarrow \dfrac{3x^{0}}{0}\) so that becomes \(\displaystyle \ln |x^{3}| + C\)
\(\displaystyle \int \dfrac{1}{x} dx \rightarrow \ln|u| + C\)
So considering \(\displaystyle \int \dfrac{3}{x} dx\) then what is\(\displaystyle u\)?
\(\displaystyle u\) must be \(\displaystyle x^{3}\) but how did it get there?
Perhaps it can be understood as
\(\displaystyle \int 3x^{-1} \rightarrow \dfrac{3x^{0}}{0}\) so that becomes \(\displaystyle \ln |x^{3}| + C\)
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I think "mad" is the wrong word here. More like frustrated. You continually post problems where you show your work (good thing), then when we critique your work, whether it's bad algebra or incorrect usage of formulas, substitution principles, etc., you tend to not correct these mistakes in your next thread post. It drives me, and I'm sure many others here, crazy to where we want to bang our heads on the wall.
Again, we really appreciate that you attempt and show all your work when you post, that is exactly what we ask for here on this site, but please heed our critiques and correct your errors fully so we aren't going round and round on these posts.
D