I'm having trouble with this integral calculus problem below. Would anyone dare attempt to solve it?

Question

ò [x (ln x)^2]^-1 dx =


these were the possible solution in the book, it may or may not help you...

a. ln |ln x| + C
b. (1/2)(ln x)2 + C
c. (3/2)(ln x)2 + C
d. - 1/|ln x| + C
e. none of these

Answer 1

∫[x (ln x)²]^-1 dx =
∫1/[x (ln x)²] dx =

u = ln x
du = dx/x
=>
∫1/[x (ln x)²] dx =∫1/u² dx = ∫u^-2 dx =
∫1/[x (ln x)²] dx = ∫u^-2 dx = -1/u + c
∫1/[x (ln x)²] dx = -1/u + c = -1/(ln x) + c

∫1/[x (ln x)²] dx = -1/(ln x) + c

Answer: letter (d)

₢

Answer 2

The correct answer is
-1 / ln(x) + C
without the absolute value enclosures. And I see that Luiz already showed you how to do the integration by parts.

Check...
d/dx {-1/ln(x)} = - d/dx {1/ln(x)}
= - d/dx {ln x}^(-1)
= - (-1) (ln x)^(-2) d/dx {ln x}
= [ 1 / (ln x)^2 ] (1/x)
= [ x (ln x)^2 ]^(-1)

Answer 3

let u=ln(x) then du=x^-1 dx so the integral becomes
int 1/u^2 du = -1/u +c = -1/ln(x) +c

(i forgot the square on the ln the first time and modified it this one's right)

Answer 4

int [dx/(x ln^2x)]

ln x = u

And dx/ x = du

int [du/u^2] = -1/u + c = - 1/lnx + C

(1/u)' = (u^-1)' = -1 u ^-2

So int (u^-2) = - 1/x

Ana

EDIT

I dont know why Doug wrote that the result is abs lnx. And Daniel solve another integral, he forgot that lnx was squared.

Answer 5

you are attempting to combine by using factors, yet that's no longer required. in case you utilize the substitution u = e^(sin(x)), then du = cos(x) e^(sin(x)) dx and the quintessential on the right part is in simple terms ? du = u + C = e^(sin(x)) + C. As to your stated therapy, no you would be unable to write ? sin(x)cos(x)e^(sin(x))dx = ? sin(x) dx ? cos(x)e^(sin(x)) dx The quintessential of the product isn't the made from the integrals. luckily, you do no longer opt for something particular because of the fact the substitution does the trick.

Answer 6

d. is the correct answer.
Note that dx/x = d Log[x], now substitute y=Log[x] and you are solving the much simpler integral dy/y^2 = -1/y.

Answer 7

FWIW, I get -1/|ln(x)|+C


Doug

Answer 8

i cant see the question properly -- what's the first symbol?

Answer 9

no way