anonymous one year ago evaluate the integral integral of (ln(4x))^2dx

1. anonymous

im assuming this is integration by parts. but i am not sure what to make my u and what to make dv

2. anonymous

would ln(4x) be good for u? and i would use chain rule to find derivative?

3. Astrophysics

$\int\limits (\ln(4x))^2 dx$ is it like this?

4. anonymous

yes it is

5. Astrophysics

You can just let u = (ln(4x))^2 and dv = 1

6. anonymous

7. anonymous

ok! where do you get the 1 from?

8. anonymous

pls help i need to get rest

9. Astrophysics

We can think of there being a 1 when we have such functions dv=dx, it's like if you have to integrate tanx dx you would use by parts, lets u = tanx, and dv = dx

10. Astrophysics

I guess it would've been better if I said dv = 1*dx

11. anonymous

ohh yes i see! thank you

12. anonymous

astro pls help

13. Astrophysics

Not tanx I meant arctanx haha

14. anonymous

hmm im not getting the right answer.. is it correct to say that the integral of ln4 is = to xln4 ?

15. anonymous

$(\ln(4x))^{2}x-\int\limits x \frac{ 2\ln4 }{ x } dx$

16. anonymous

so then i cancel out the x on the outside and the x in the denominator..

17. anonymous

and i am left with the integral of 2ln4, so then i take out the 2 and put it behind the integral, and i am left with the integral of ln4

18. anonymous

from there i took the integral of ln4 which i believe is xln4 and i put both parts together but the answer is incorrect

19. anonymous

$(\ln(4x))^2x-2xln4$ that is my answer..

20. anonymous

21. imqwerty

the answer is $x(\ln (4x))^2-2xln (4x)+2x$ recheck once again :)

22. anonymous

You're almost there.

23. anonymous

thank you.. i am not seeing where the 2x at the end comes from.. ?

24. anonymous

25. anonymous

that is my work so far

26. imqwerty

u=4x $\int\limits\frac{ (\ln (u)) }{ 4}du$after some vry bad calculations nd simplifications u get- $\frac{ 1 }{ 4 }((u)(\ln (u))^2 -2\ln (u)+2(u))$and then when u simplify nd put u=4x then u get that answer..

27. anonymous

ok i see, thank you!

28. imqwerty

no prblm but that is a long method..