anonymous one year ago 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.

1. 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 :)

2. 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.

3. 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: $2 \int \ln (t) dt$

4. 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

5. Empty

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

6. Empty

You have to use integration by parts like I described earlier

7. Empty

Here, to show it's not the integral, then this should be true if you want to check it $\frac{d}{dx}( x \ln(-x)) \ne \ln x$

8. 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.

9. Empty

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

10. anonymous

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

11. 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

12. 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

13. 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.

14. 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 $I=\frac{1}{2}.\int\limits \ln(t).1dt$

15. anonymous

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

16. 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 $t=ax+b$