integrate (ln(x)*(1/x))dx?

Question

Using Integration by Parts...
Mine just seemed to go in circles..please help.

Answer 1

it does not need integration by parts sience the derivetive of ln(x) is 1/x then the answer canbe easily found 1/2*(ln(x))^2 you can test it ;)

Answer 2

You don't really need integration by parts, because this is a function (ln x) multiplied by the derivative(1/x)
Use a simple substitution u= ln x
then du/dx = 1/x
which means 1/x dx can be replaced by du and ln x is replaced by u
then its integral u du = u^2/2 +c = (lnx)^2 / 2 +c

with integration by parts
integral, I = ln x * ln x - integral ln x * 1/x dx
then integral = (ln x)^2 - integral
so 2 * integral = (ln x)^2
integral = (ln x)^2 / 2
Here we can just move integrals around only because the same integral arose from the f(x) * f'(x) dx form the original integral possesses.

Answer 3

There's no need to use integration by parts because the derivative of ln(x) is 1/xdx. Using the integration formula
u^ndu=u^n+1/n+1, let u=ln(x) du=1/xdx n=1
final ans. is 1/2(ln(x))^2 + c

Answer 4

should be 1/2(ln x )^2 + C

Answer 5

(ln(x) * (1/x)) dx = ln(x) d(ln(x)) .
Intereration will be c + (ln(x))^2

Answer 6

put ln x=t
1/x*dx=dt
int tdt=t^2/2+C=(lnx)^2+C

Answer 7

Ln x---------------------(1)function
1/x-----------------------(2)function
ans=lnx*lnx-integration((1/x)*lnx)=lnx*lnx - ans
=> 2*ans =lnx * lnx
=>ans=(lnx *lnx) /2 + c
hope u understand the ans