## 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 Group Title

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

3. SheldonEinstein Group Title

May I help?

4. satellite73 Group Title

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 Group Title

(x)^4 ?

7. satellite73 Group Title

are you familiar with "u - substitution"

8. SheldonEinstein Group Title

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

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

10. satellite73 Group Title

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 Group Title

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 Group Title

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

14. satellite73 Group Title

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

yes @SheldonEinstein

16. SheldonEinstein Group Title

I would not interupt and let @satellite73 answer.

17. satellite73 Group Title

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 Group Title

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 Group Title

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

Ok I'm with you so far

21. satellite73 Group Title

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

22. satellite73 Group Title

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

why over 5? that's where I got lost

24. satellite73 Group Title

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

25. satellite73 Group Title

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 Group Title

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

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

29. satellite73 Group Title

ok

30. satellite73 Group Title

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 Group Title

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 Group Title

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 Group Title

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

35. SheldonEinstein Group Title

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