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.

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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
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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 :)
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.
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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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
The integral of ln x is not xln(-x)
You have to use integration by parts like I described earlier
Here, to show it's not the integral, then this should be true if you want to check it \[\frac{d}{dx}( x \ln(-x)) \ne \ln x\]
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.
No, that's not right either be careful! \[\int \frac{1}{x}dx = \ln x\] \[\int \ln x dx \ne \frac{1}{x}\]
\[\int\limits \ln(x)dx=x.\ln(x)-x+C\]
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
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
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.
\[t=2x+3\]\[\frac{dt}{dx}=2\] \[dt=2dx\]\[\implies dx=\frac{dt}{2}\] \[I=\frac{1}{2}\int\limits \ln(t)dt\] Integrate by parts, where your first function is ln(t) and second function is constant function 1 \[I=\frac{1}{2}.\int\limits \ln(t).1dt\]
You'll arrive at the fact that \[\int\limits \ln(t)dt=t.\ln(t)-t+C\]
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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