This discussion is now closed.

Check out other Related discussions

I don't understand WHY the integral of 1/(xlnx) is ln(lnx) + C

I need to integrate 1/(xlnx)

I'm doing so by integration by parts

1/(xlnx) = 1/x . 1/ lnx

u = 1/lnx dv/dx = 1/x

I integrate and use uv - int(du/dx . v) to get 1 which if definitely wrong

I've looked at how people do it over the net, and they always take u as lnx?
I cannot for the life of me work out why?

When we integrate by parts we: int(u.dv/dx)
why isn't it working for me?

My u value is 1/lnx and my dv/dx is 1/x Becuase when multiply these two together we get 1/(xlnx)

(edited 12 years ago)

Reply 1

claret_n_blue

You can't integrate ln(x), you can only differentiate it to 1/x so therefore you need to set u = ln(x) so you can differentiate it.

Reply 2

Sifr

Original post by claret_n_blue

You can't integrate ln(x), you can only differentiate it to 1/x so therefore you need to set u = ln(x) so you can differentiate it.

Yes I know that's why I chose u= 1/ lnx ...
Just don't get why everyone gets u = lnx :/

Reply 3

f1mad

Original post by Sifr

Yes I know that's why I chose u= 1/ lnx ...
Just don't get why everyone gets u = lnx :/

Because its differential is 1/x, and you already have 1/x in the integrand (for substitution that is).

(edited 12 years ago)

Reply 4

porkstein

You choose u = lnx so that dx = xdu. This then cancels with the other x on the bottom of the fraction.

Reply 5

Chwirkytheappleboy

Do you have to use integration by parts?

\dfrac{1}{x\ln{x}}=\dfrac{\frac{1}{x}}{\ln{x}}

What is

\displaystyle \int \dfrac{f'(x)}{f(x)}

Reply 6

f1mad

Original post by Chwirkytheappleboy

Do you have to use integration by parts?

\dfrac{1}{x\ln{x}}=\dfrac{\frac{1}{x}}{\ln{x}}

What is

\displaystyle \int \dfrac{f'(x)}{f(x)}

No you don't; you could use what you have done or substitution too.

Reply 7

claret_n_blue

Original post by Sifr

Yes I know that's why I chose u= 1/ lnx ...
Just don't get why everyone gets u = lnx :/

Ohh. Well if you differentiate it you get 1/x and this can be substituted for the 1/x. You don't need to use by parts, you use substitution:

\int \frac{1}{x \cdot ln(x)} dx

Let u = ln(x)
du = (1/x) dx

\int \frac{1}{x \cdot u} dx = \int \frac{1}{u} du = ln(u) + C = ln (ln x) + C

Reply 8

Sifr

Original post by Chwirkytheappleboy

Do you have to use integration by parts?

\dfrac{1}{x\ln{x}}=\dfrac{\frac{1}{x}}{\ln{x}}

What is

\displaystyle \int \dfrac{f'(x)}{f(x)}

I can use another method, but it's the principle, every method should work? I'm really pedantic and it bugs me

Reply 9

f1mad

Original post by Sifr

I can use another method, but it's the principle, every method should work? I'm really pedantic and it bugs me

Not necessarily, that's why we have various integration methods.

Consider trying to integrate ln(x).

This isn't possible via substitution, only by parts.

Reply 10

porkstein

Well perhaps you could show the working that gave you the answer 1? Because that's not what I get when I do it by parts...

Reply 11

Chwirkytheappleboy

Original post by Sifr

I can use another method, but it's the principle, every method should work? I'm really pedantic and it bugs me

So if you're saying

u = \dfrac{1}{\ln{x}}, \dfrac{dv}{dx} = \dfrac{1}{x}

then we have

\dfrac{du}{dx} = -\dfrac{1}{x(\ln{x})^2}

and

v = \ln{x}

\displaystyle \int \dfrac{1}{x\ln{x}} = 1 + \displaystyle \int \dfrac{1}{x\ln{x}} + c

Since the second term of the right hand side is the same as the left hand side, we end up with the wonderous result of -c = 1. This is pretty useless, so since it doesn't get us anywhere we say "ok let's try another method". Use the easier way I suggested earlier and then go to bed

Reply 12

SZRoberson

IBP (Integration by Parts) is not necessary in this problem. I encountered this integral in a homework site a long time ago.

Let's use substitution.

\displaystyle\int \frac {1}{x\ln x}\ dx

is our integral. I see I have a function (the natural logarithm) and its derivative

Unparseable latex formula:

\left \dfrac {1}{x}\right

sitting in the integrand. I'll make the following substitution:

u = ln\ x[br]du = \dfrac {1}{x}\ dx

Our integrand now becomes

\displaystyle\int \frac {1}{u}\ du

. This we can do easily, and you should grab

ln (ln\ x) + C

Reply 13

WTSFG

Cos mama said, mama said that if you take it and integrate it, that's what you get

Reply 14

nuodai

Original post by Sifr

I can use another method, but it's the principle, every method should work? I'm really pedantic and it bugs me

There's no reason to think that integration by parts should work just because you have a product in your integral. Often the best way to deal with a product is integration by substitution, particularly when one of the terms is a function of a function and the other term is the derivative of the 'inside function'.

Think about it: integration by parts states

\displaystyle \int f(x)\cdot \dfrac{dg}{dx}\, dx = f(x)g(x) - \int \dfrac{df}{dx} \cdot g(x)\, dx

Compare this with the product rule (rearranged):

f(x) \cdot \dfrac{dg}{dx} = \dfrac{d}{dx}(f(x)g(x)) - \dfrac{df}{dx} \cdot g(x)

And integration by substitution states

\displaystyle \int f(u(x))\dfrac{du}{dx}\, dx = \int f(u)\, du

Compare this with the chain rule:

\dfrac{df}{du} \dfrac{du}{dx} = \dfrac{d}{dx} f(u(x))

As you can see, integration by parts reverses the product rule (which you can think of as 'differentiation by parts') and integration by substitution reverses the chain rule (which you can think of as 'differentiation by substitution').

Both LHS integrals involve products, but what you get out is a very different type of function. Only in special rare cases are both methods appropriate for calculating the antiderivative.

(edited 12 years ago)

Reply 15

TheJ0ker

Original post by Chwirkytheappleboy

Do you have to use integration by parts?

\dfrac{1}{x\ln{x}}=\dfrac{\frac{1}{x}}{\ln{x}}

What is

\displaystyle \int \dfrac{f'(x)}{f(x)}

This is the easiest way to do it, writing it like that shows the integral is clearly ln(ln(x)) + c

Reply 16

Philor9294

Not sure if you still need an answer, but I got 1/2 x ln(x)² + c

used integrating by parts.

u=lnx u'=1/x
v'=1/x v=lnx

integral(1/xlnx)dx = (lnx)² - integral(1/xlnx)dx
2integral(1/xlnx)dx = (lnx)²
integral (1/xlnx)dx = 1/2 x (lnx)² + c