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.

Question

empty · Answer

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 · Answer

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 · Answer

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:

$$2 \int \ln (t) dt$$

anonymous · Answer

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 · Answer

The integral of ln x is not xln(-x)

empty · Answer

You have to use integration by parts like I described earlier

empty · Answer

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 · Answer

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 · Answer

No, that's not right either be careful!

$$\int \frac{1}{x}dx = \ln x$$
$$\int \ln x dx \ne \frac{1}{x}$$

anonymous · Answer

$$\int\limits \ln(x)dx=x.\ln(x)-x+C$$

anonymous · Answer

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 · Answer

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 · Answer

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 · Answer

$$t=2x+3$$$$\frac{dt}{dx}=2$$
$$dt=2dx$$$$\implies dx=\frac{dt}{2}$$
$$I=\frac{1}{2}\int\limits \ln(t)dt$$
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 · Answer

You'll arrive at the fact that
$$\int\limits \ln(t)dt=t.\ln(t)-t+C$$

anonymous · Answer

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
$$t=ax+b$$