anonymous
  • anonymous
How do I integrate the term int[ln(2x+3)dx]? I know the answer will be (x-3/2)ln(2x-3), but I don't know how to get there.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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Empty
  • Empty
You have to use a clever bit of integration by parts. Choose: \(u=\ln(2x+3)\) and \(dv=dx\) this will give you: \(du = \frac{2}{2x+3}dx\) and \(v=x\) So we can get: \[\int \ln(2x+3) dx = x\ln(2x+3)-\int \frac{2x}{2x+3}dx\] Then continue solving :)
anonymous
  • anonymous
I have to do it by substitution, and the (2x+3) has to be what's subsituted. We use t as substition where i'm from, so right now i'm at \[\int\limits_{?}^{?}\ln(2x-3)dx = \int\limits_{?}^{?}\ln(t) \] where \[t=2x+3 \] and \[t'=2\] I don't know how to proceed from there.
Empty
  • Empty
Well this will help simplify it but you won't be able to solve it unless you've memorized \(\int \ln(x) dx\). So by substitution following as you're doing: \[\int \ln(2x+3)dx\] Now we take \[t=2x+3\] \[\frac{dx}{dt}=2\] Multiply both sides of this by dt to get: \[dx=2dt\] So now we can plug it back in: \[\int \ln(2x+3)dx = \int \ln (t) 2dt\] Then we can pull the constant 2 out: \[2 \int \ln (t) dt\]

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anonymous
  • anonymous
So if we say \[2\int\limits_{}^{}\ln(t)dt \] and you then integrate ln, then that must eqaul to \[2(x*\ln-x)\] And I can't see how that can be \[(x-\frac{ 3 }{ 2 })\ln(2x-3) \] which is the answer
Empty
  • Empty
The integral of ln x is not xln(-x)
Empty
  • Empty
You have to use integration by parts like I described earlier
Empty
  • Empty
Here, to show it's not the integral, then this should be true if you want to check it \[\frac{d}{dx}( x \ln(-x)) \ne \ln x\]
anonymous
  • anonymous
Oh, damn you're right. Thanks! So the integral to \[\int\limits_{}^{}\ln = \frac{ 1 }{ x }\] It's slowly making sense :) except can you explain to me why it's \[\frac{ 2 }{ 2x+3 }\] I don't understand what the 3 is doing there.
Empty
  • Empty
No, that's not right either be careful! \[\int \frac{1}{x}dx = \ln x\] \[\int \ln x dx \ne \frac{1}{x}\]
anonymous
  • anonymous
\[\int\limits \ln(x)dx=x.\ln(x)-x+C\]
anonymous
  • anonymous
That leads me literally no where. so if \[\int\limits_{}^{}\ln(t) = t*\ln(t)-t \] then it's \[2x-3*\ln(2x-3)-2x+3 \] Which is nowhere near what the end result has to be
Empty
  • Empty
Last thing I'll say, the correct answer is really \[(x+\frac{3}{2})(\ln(2x+3)-1)\] You have all the information you need in this thread, you just need to look through it and see if you made any tiny errors along the way. http://www.wolframalpha.com/input/?i=int%20ln(2x%2B3)dx&t=crmtb01
anonymous
  • anonymous
You can also work this in a somewhat inefficient, roundabout way with a substitution (but you can't avoid IBP, I'm afraid). Setting \(e^u=2x+3\) gives \(\dfrac{1}{2}e^u\,du=dx\), making the integral equivalent to \[\frac{1}{2}\int e^u\ln e^u\,du=\frac{1}{2}\int ue^u\,du\] I've made a similar comment on another question the other day mentioning that it might be easier to recognize which parts are to be designated \(f\) and \(dg\) in the IBP formula.
anonymous
  • anonymous
\[t=2x+3\]\[\frac{dt}{dx}=2\] \[dt=2dx\]\[\implies dx=\frac{dt}{2}\] \[I=\frac{1}{2}\int\limits \ln(t)dt\] Integrate by parts, where your first function is ln(t) and second function is constant function 1 \[I=\frac{1}{2}.\int\limits \ln(t).1dt\]
anonymous
  • anonymous
You'll arrive at the fact that \[\int\limits \ln(t)dt=t.\ln(t)-t+C\]
anonymous
  • anonymous
In general, for a function \[f(ax+b)\] We have \[\int\limits f(ax+b)dx=\frac{1}{a} \int\limits f(ax+b)d(ax+b)=\frac{1}{a}\int\limits f(t)dt\] where \[t=ax+b\]

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