Quick on recognizing Integration by Parts

FritoTaco

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Hello,

I just need quick help on recognizing why we need to use Integration by Parts. I know it's when there's a product of two different functions when integrating, but what is the two different functions in this equation?

\(\displaystyle \int k\ln(t)dt\)
\(\displaystyle k\int \ln(t)dt\) Factored K outside integrand

Letter 'k' is a constant in this case. So how do we recognize to use Integration by Parts here? \(\displaystyle Ln\) is natural log which is one function, is the other function the \(\displaystyle t\) inside the natural log? Making it a product of two different functions? If this is the case, would we let \(\displaystyle u = \ln(t)\), then \(\displaystyle du = \dfrac{1}{t} dt\) and \(\displaystyle dv = 1dt\) and \(\displaystyle v = t\)




Thank you.
 
Hello,

I just need quick help on recognizing why we need to use Integration by Parts. I know it's when there's a product of two different functions when integrating, but what is the two different functions in this equation?

\(\displaystyle \int k\ln(t)dt\)
\(\displaystyle k\int \ln(t)dt\) Factored K outside integrand

Letter 'k' is a constant in this case. So how do we recognize to use Integration by Parts here? \(\displaystyle Ln\) is natural log which is one function, is the other function the \(\displaystyle t\) inside the natural log? Making it a product of two different functions? If this is the case, would we let \(\displaystyle u = \ln(t)\), then \(\displaystyle du = \dfrac{1}{t} dt\) and \(\displaystyle dv = 1dt\) and \(\displaystyle v = t\)

Thank you.

Yes, your reasoning is pretty much exactly correct. Any function f(x) can be written as the product of f(x) and 1. In most cases, this will produce nothing meaningful, but in some cases it can simpify the integral, as is the case here. If we follow through with your proposed solution and use the integration by parts formula, we find:

\(\displaystyle \displaystyle \int u \: dv = uv - \int v \: du \implies \int \ln(t) \: dt = \ln(t) \cdot t - \int t \cdot \frac{1}{t} \: dt\)

Can you finish up from here? What do you get at the very end? If you differentiate your answer, do you get ln(t)?
 
Yes, your reasoning is pretty much exactly correct. Any function f(x) can be written as the product of f(x) and 1. In most cases, this will produce nothing meaningful, but in some cases it can simpify the integral, as is the case here. If we follow through with your proposed solution and use the integration by parts formula, we find:

\(\displaystyle \displaystyle \int u \: dv = uv - \int v \: du \implies \int \ln(t) \: dt = \ln(t) \cdot t - \int t \cdot \frac{1}{t} \: dt\)

Can you finish up from here? What do you get at the very end? If you differentiate your answer, do you get ln(t)?
Yeah, I got what you have here and then I simplified it.



\(\displaystyle \displaystyle \int u \: dv = uv - \int v \: du \implies \int \ln(t) \: dt = \ln(t) \cdot t - \int t \cdot \frac{1}{t} \: dt\)

\(\displaystyle \displaystyle \int \ln(t)dt = \ln(t)\cdot t - \int 1 dt\)

\(\displaystyle \displaystyle \int \ln(t)dt = t\ln(t) - t\)

Hopefully, this is correct.
 
Yeah, I got what you have here and then I simplified it.

\(\displaystyle \displaystyle \int u \: dv = uv - \int v \: du \implies \int \ln(t) \: dt = \ln(t) \cdot t - \int t \cdot \frac{1}{t} \: dt\)

\(\displaystyle \displaystyle \int \ln(t)dt = \ln(t)\cdot t - \int 1 dt\)

\(\displaystyle \displaystyle \int \ln(t)dt = t\ln(t) - t\)

Hopefully, this is correct.

Well, as I suggested previously, if you're ever unsure of an answer, you can always check it yourself. For this problem, what did you get when you took the derivative of your answer? As you (should) know, derivation and integration are inverse functions. So, if you end up with ln(t), then you know you did it right. If not, then you can begin to search for any error(s) you may have made. As a technical note, the actual solution involves a constant of integration, but that will disappear when you take the derivative.
 
Well, as I suggested previously, if you're ever unsure of an answer, you can always check it yourself. For this problem, what did you get when you took the derivative of your answer? As you (should) know, derivation and integration are inverse functions. So, if you end up with ln(t), then you know you did it right. If not, then you can begin to search for any error(s) you may have made. As a technical note, the actual solution involves a constant of integration, but that will disappear when you take the derivative.

I took the antiderivative of 1dt to just get t. I did not get ln(t) for my answer and I looked, but cannot find any mistakes?
 
I took the antiderivative of 1dt to just get t. I did not get ln(t) for my answer and I looked, but cannot find any mistakes?

Yes, this is a mathematically valid answer, but it's the answer to the wrong question. The question you're trying to answer is "What is the integral of ln(t) with respect to t?" to which your answer is not 1, but instead \(\displaystyle t \cdot \ln(t) - t\). If you derive that with respect to t, you'll find you do, indeed, get ln(t) ;)
 
..., the actual solution involves a > > > constant of integration < < < , but that will disappear when you take the derivative.
\(\displaystyle \displaystyle \int \ln(t)dt = t\ln(t) - t\)

Hopefully, this is correct.


FritoTaco, write that inclusion of that constant of integration like this:

\(\displaystyle \displaystyle \int \ln(t)dt = t\ln(t) - t + C\)
 
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