heradog Group Title I need help with this step by step Evaluate the indefinite integral. \int \frac{(\ln(x))^4}{x} dx by substitution one year ago one year ago

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

2. satellite73

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

3. SheldonEinstein

May I help?

4. satellite73

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

pretend I'm a complete moron

6. SheldonEinstein

(x)^4 ?

7. satellite73

are you familiar with "u - substitution"

8. SheldonEinstein

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

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

10. satellite73

it is the chain rule backwards

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. satellite73

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. SheldonEinstein

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

14. satellite73

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

yes @SheldonEinstein

16. SheldonEinstein

I would not interupt and let @satellite73 answer.

17. satellite73

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. satellite73

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. satellite73

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

Ok I'm with you so far

21. satellite73

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

22. satellite73

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

why over 5? that's where I got lost

24. satellite73

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

25. satellite73

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

nevermind

27. satellite73

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

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

29. satellite73

ok

30. satellite73

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. SheldonEinstein

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. satellite73

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

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

34. satellite73

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

35. SheldonEinstein

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