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anonymous
 one year ago
How do I integrate the term int[ln(2x+3)dx]? I know the answer will be (x3/2)ln(2x3), but I don't know how to get there.
anonymous
 one year ago
How do I integrate the term int[ln(2x+3)dx]? I know the answer will be (x3/2)ln(2x3), but I don't know how to get there.

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Empty
 one year ago
Best ResponseYou've already chosen the best response.4You 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
 one year ago
Best ResponseYou've already chosen the best response.0I 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 \[\int\limits_{?}^{?}\ln(2x3)dx = \int\limits_{?}^{?}\ln(t) \] where \[t=2x+3 \] and \[t'=2\] I don't know how to proceed from there.

Empty
 one year ago
Best ResponseYou've already chosen the best response.4Well 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: \[2 \int \ln (t) dt\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So if we say \[2\int\limits_{}^{}\ln(t)dt \] and you then integrate ln, then that must eqaul to \[2(x*\lnx)\] And I can't see how that can be \[(x\frac{ 3 }{ 2 })\ln(2x3) \] which is the answer

Empty
 one year ago
Best ResponseYou've already chosen the best response.4The integral of ln x is not xln(x)

Empty
 one year ago
Best ResponseYou've already chosen the best response.4You have to use integration by parts like I described earlier

Empty
 one year ago
Best ResponseYou've already chosen the best response.4Here, 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
 one year ago
Best ResponseYou've already chosen the best response.0Oh, 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
 one year ago
Best ResponseYou've already chosen the best response.4No, that's not right either be careful! \[\int \frac{1}{x}dx = \ln x\] \[\int \ln x dx \ne \frac{1}{x}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits \ln(x)dx=x.\ln(x)x+C\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That leads me literally no where. so if \[\int\limits_{}^{}\ln(t) = t*\ln(t)t \] then it's \[2x3*\ln(2x3)2x+3 \] Which is nowhere near what the end result has to be

Empty
 one year ago
Best ResponseYou've already chosen the best response.4Last 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
 one year ago
Best ResponseYou've already chosen the best response.0You 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
 one year ago
Best ResponseYou've already chosen the best response.0\[t=2x+3\]\[\frac{dt}{dx}=2\] \[dt=2dx\]\[\implies dx=\frac{dt}{2}\] \[I=\frac{1}{2}\int\limits \ln(t)dt\] 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
 one year ago
Best ResponseYou've already chosen the best response.0You'll arrive at the fact that \[\int\limits \ln(t)dt=t.\ln(t)t+C\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0In 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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