## anonymous 3 years ago I need help with this step by step Evaluate the indefinite integral. \int \frac{(\ln(x))^4}{x} dx by substitution

1. anonymous

$\int\limits_{}^{}\frac{ \ln(x)^4 }{ x }dx$

2. anonymous

try $$u=ln(x), du =\frac{dx}{x}$$ and you should get it in one step

3. anonymous

May I help?

4. anonymous

visualize it as $\int\frac{ \ln(x)^4 }{ x }dx=\int \ln^4(x)\frac{dx}{x}$

5. anonymous

pretend I'm a complete moron

6. anonymous

(x)^4 ?

7. anonymous

are you familiar with "u - substitution"

8. anonymous

$$\ln x^4$$ Why $$\ln ^4 x$$ @satellite73

9. anonymous

sorry, it's the whole thing to the 4th

10. anonymous

it is the chain rule backwards

11. anonymous

I understand that you pick one component to make u and that the derivative of that needs to be in the function it's what you do with it that I can't figure out

12. anonymous

first off, lets get the answer and see why it works. what do you get when you take the derivative of $$\ln^3(x)$$ ? by the chain rule you get $$3\ln(x)\times \frac{1}{x}=\frac{3\ln^2(x)}{x}$$

13. anonymous

Ok! $$(\ln x)^4$$ is what we have

14. anonymous

this tells you $\int\frac{3\ln^2(x)}{x}dx=\ln^3(x)$

15. anonymous

yes @SheldonEinstein

16. anonymous

I would not interupt and let @satellite73 answer.

17. anonymous

so a u-sub is a way to undo the chain rule you have a composite function $$\ln^4(x)=(\ln(x))^4$$ multiplied by the derivative of the "inside function" $$\frac{1}{x}$$

18. anonymous

if you make the substitution $$u=\ln(x)$$ then the derivative is $$\frac{1}{x}$$ so the gimmick is to say $u=\ln(x), du=\frac{1}{x}dx$

19. anonymous

then rewrite with only $$u$$ terms and get $\int u^4du$

20. anonymous

Ok I'm with you so far

21. anonymous

now the anti derivative is relatively easy it is $$\frac{u^5}{5}$$

22. anonymous

putting the $$u=\ln(x)$$ back we get $\frac{\ln^5(x)}{5}$

23. anonymous

why over 5? that's where I got lost

24. anonymous

check that this works by differentiation, and you will see that this is correct

25. anonymous

oh, over 5 because the derivative of $$u^5$$ is $$5u^4$$ but you want $$u^4$$ and not $$5u^4$$ so you have to divide by $$5$$ to get it

26. anonymous

nevermind

27. anonymous

$\int x^n dx =\frac{x^{n+1}}{n+1}$

28. anonymous

I'm used to doing (1/5)u^5

29. anonymous

ok

30. anonymous

check that the derivative of $\frac{1}{5}\ln^5(x)$ is $\frac{\ln^4(x)}{x}$ and you will see why the "u - substitution" works

31. anonymous

I did like this : $\int \cfrac{ [\ln x]^4 dx }{x}$ Put $$\cfrac{1}{5} [\ln x]^5 = u$$ $$\int du = u + c$$ = $$\cfrac{1}{5} [\ln x]^5 + C$$ $$\implies \cfrac{[\ln x]^4 dx}{x} = du$$ Is this also correct @satellite73 ?

32. anonymous

it looks like you knew the answer at the start and then just said it look at this line $\int du = u + c$ this says the integral is $$u$$ which you knew at the beginning, probably by doing the u - sub in your head in other words, you knew what the answer was, but the u - sub is the way to get the answer

33. anonymous

both of those really help! I'll try the next one and see what happens

34. anonymous

the entire problem is to figure out the $\frac{1}{5}\ln^5(x)$ part

35. anonymous

Ok! I got it. @heradog not me, @satellite73 was the one who helped. Best of luck for your journey and btw $$\mathbb{ I} \space \mathbb{LOVE} \space \mathbb{INTEGRALS}$$

36. anonymous

good i'm stuck again