# How do you integrate this?

1. Sep 24, 2009

### -EquinoX-

1. The problem statement, all variables and given/known data

$$\int (3x-5)e^{4ln(x)}$$

2. Relevant equations

3. The attempt at a solution

Any ideas?

2. Sep 24, 2009

### aPhilosopher

rewrite $$e^{4ln(x)}$$ as $$e^{(ln x)4}$$ and remember your rules for exponents!

3. Sep 25, 2009

### -EquinoX-

does e^(ln(x)) becomes x?

4. Sep 25, 2009

### aPhilosopher

yes.

5. Sep 25, 2009

### -EquinoX-

http://img85.imageshack.us/img85/8667/questionl.th.jpg [Broken]

If the original question is above and I solved the integral above which results in:

$$x^3 - \frac{5x^2}{2} + \frac{C}{e^{4ln(x)}$$

then I need to divide this by $$e^{4ln(x)}$$ right??
I am also given an initial state of y(1) = 5

the final result that I get is :

2x^2-5x+13/x

why is this wrong? which step am I not doing it right?

however web assign marks it as wrong.. am I doing something wrong?

Last edited by a moderator: May 4, 2017
6. Sep 25, 2009

### aPhilosopher

Apparently.

What's $$\int_0^x\frac{4}{t}dt$$

7. Sep 25, 2009

### Eagle_Eng

Easy mate
integration of (3x-5) = 3x^2-5x
hence integration of e^4ln(x) is = e nd ln cancell each other log rules..
hence take 4 to the other side becomes power. = x^4
final ans
(3x^2 - 5x).(x^4)...so u can do the rest i assume.

to -EquinoX

8. Sep 25, 2009

### -EquinoX-

well $$e^{4ln(x)}$$ is equal to $$e^4x$$ right?

if that's so the the integration becomes $$e^4 \int (3x^2-5x)$$

which is $$(x^3 - \frac{5x^2}{2})e^4 + C$$

is this correct?

if so then I divide $$(x^3 - \frac{5x^2}{2})e^4 + C$$ by $$e^{4ln(x)}$$ and then simplifying that I get $$(x^2 - \frac{5x}{2}) + \frac{C}{e^{4ln(x)}}$$

Last edited: Sep 25, 2009
9. Sep 25, 2009

### aPhilosopher

$$e^{4 + ln x} = e^{4}e^{ln x} = e^{4}x$$
$$e^{4ln x} = e^{(ln x)4} = (e^{ln x})^{4} = x^{4}$$

I thought we already went over this. As to the integral question I asked earlier, what's ln 0?

10. Sep 25, 2009

### -EquinoX-

shouldn't it be the integral of (3x-5)*x^4\

I am just confused why you integrate (3x-5) first and then e^4ln(x) separately... as I recall they were multiplied

11. Sep 25, 2009

### aPhilosopher

I'm not doing that. I'm finding the integrating factor. That's how you got $$e^{4ln x}$$, right? By performing the integral I gave above?

12. Sep 25, 2009

### -EquinoX-

okay so the integral should be $$\int 3x^5-5x^4 = \frac{x^6}{2} - x^5 + C$$ then divide all of this by $$x^4$$ (integrating factor) I will get $$\frac{x^2}{2} - x + \frac{C}{x^4}$$

put in the initial condition which is y(1) = 5,. I can solve for C which is 11/2 then plug it back in I will have $$y = \frac{x^2}{2} - x + \frac{11}{2x^4}$$ as a particular solution, true or not?

Last edited: Sep 25, 2009
13. Sep 25, 2009

### aPhilosopher

Yes, but in the future, once you have the solution, you can always check it for yourself.

14. Sep 25, 2009

### -EquinoX-

and once again reason why I post it is because web assign doesn't accept that answer, I re did the problem couple of times and it goes down to that answer... I hate this thing

15. Sep 25, 2009

### aPhilosopher

That's harsh. I hated those things when I had to do them. It's right though. I plugged in for 1 and plugged the solution into the ODE you gave. They're both right so it's not your fault. Maybe put it all in one fraction or something.

16. Sep 25, 2009

### -EquinoX-

okay...I'll see what I can do...