integrate (lnx)^2

How on earth do you integrate (ln x)^2

Iv tried by parts but whatever happens, I get back to needing to integrating ln x.

What iv done is:

let u = ln x let lnx dx = dv
du = 1/x dx xlnx-x = v

therefore:

(xlnx-x)(lnx) - (integral) xlnx -x.1/x dx

and then i get stuck lol

Havent got a clue, any help much appreciated!

Reply 1

inabit

i was gonna do this question earlier - but gave up....

lol

then parts again!

OP

lol i was dreading for you to say that.. such a headwreck this question lol

Reply 4

Spire

Would have suggested by parts, but thinking about it now I don't think its possible that way. Only thing that pops into my head right now is an approximation using infinite/maclaurin series possibly.

Reply 5

dietwater

OP

2

how would i integrate xlnx - 1 by parts though? let u=xlnx?

Would that just differentiate to 1?

Reply 6

dietwater

OP

2

Spire

Would have suggested by parts, but thinking about it now I don't think its possible that way. Only thing that pops into my head right now is an approximation using infinite/maclaurin series possibly.

O.o im lost lol, we dont cover that in our course..

Reply 7

Scooby1964

6

Nah, there's an easier way.

Make u (lnx)^2 and dv/dx 1

So u=(ln x)^2 and v=x

Integrate by parts, and you should get x(ln x)^2 - 2x ln x + 2x + C

Heinemann Core Maths 4 for Edexcel? Page 102 ex. 6G q. 2d? Me too! :smile:

Reply 8

Dadeyemi

13

u=lnx
du=dx/x
=>dx=xdu=e^u.du
(lnx)^2.dx= u^2.e^u.du

Then parts many times does it I believe.

Reply 9

madima

dietwater

how would i integrate xlnx - 1 by parts though? let u=xlnx?

Would that just differentiate to 1?

u = ln x, dv/dx = x; -1 just integrates to -x.

Reply 10

Scooby1964

6

dietwater

how would i integrate xlnx - 1 by parts though? let u=xlnx?

Would that just differentiate to 1?

Nooooo nonono...

You can only have u equal to ONE term when you integrate by parts. You'd need to integrate the x ln x using int. by parts (u = ln x, dv/dx = x) and integrate the 1 separately after.

See my post above - hope it helps :wink:

Reply 11

dietwater

OP

2

Scooby1964

Nah, there's an easier way.

Make u (lnx)^2 and dv/dx 1

So u=(ln x)^2 and v=x

Integrate by parts, and you should get x(ln x)^2 - 2x ln x + 2x + C

Heinemann Core Maths 4 for Edexcel? Page 102 ex. 6G q. 2d? Me too! :smile:

How do you get du? how do u differentiate (lnx)^2

Reply 12

Scooby1964

6

dietwater

How do you get du? how do u differentiate (lnx)^2

Use the product rule.

The result will be in the form u(dv/dx)+v(du/dx). (sorry, that's standard form, they use u and v for it. Since this is all taking place within the "u" IBP term, maybe call them a and b for clarity when differentiating).

So a=ln x and b=ln x

da/dx=1/x and db/dx = 1/x

So differentiating we get (ln x)/x + (ln x)/x = 2(ln x)/x

See it? If not, quote me

Reply 13

Adjective

12

dietwater

How do you get du? how do u differentiate (lnx)^2

Chain rule, u = ln(x).

Reply 14

Adjective

12

Scooby1964

Use the chain rule.

The result will be in the form u(dv/dx)+v(du/dx). (sorry, that's standard form, they use u and v for it. Since this is all taking place within the "u" IBP term, maybe call them a and b for clarity when differentiating).

So a=ln x and b=ln x

da/dx=1/x and db/dx = 1/x

So differentiating we get (ln x)/x + (ln x)/x = 2(ln x)/x

See it? If not, quote me

Actually, that's the product rule, but it works just as well! :wink:

Reply 15

Scooby1964

6

Patent Pending

Actually, that's the product rule, but it works just as well! :wink:

Damn - now who's the prize idiot? :s-smilie:

Reply 16

Totally Tom

18

I get

x(lnx)^2-2xlnx+2lnx+c

anyone agree?

hmm

this doesn't differentiate back nicely.

checking workings...

Reply 17

The Lone Wanderer

Let

u=lnx

. Hence

\frac {du}{dx}=\frac{1}{x}

and also

x=e^u

.
Hence,

\int \ (lnx)^2\, dx = \int \ u^2 x\,du = \int \ u^2e^u\, du

and from here calculate the integral using parts.
I am not quite sure of this, but it seems to be working...

Reply 18

Scooby1964

6

Totally Tom

I get

(lnx)^2-2xlnx+2lnx+c

anyone agree?

Nope, I diagree (I've got the advantage of having the book though, so I can see the answer :wink:

)

\int(ln x)^2 dx

Let u=

(ln x)^2

so du/dx=2(ln x)/x

Let v=x so dv/dx=1

So

\int(ln x)^2 dx=x(ln x)^2-\int2x\frac{ln x}{x}dx

=x (ln x)^2 - INT[2 ln x] dx

Use IBP again

Let a=ln x so da/dx=1/x

Let b=2x so db/dx=2

So integrating gives 2x ln x - INT[2]dx = 2x ln x - 2x

Put into original integral

=x(ln x)^2-[2x ln x - 2x]+C

=x(ln x)^2 - 2x ln x + 2x + C

Hope this helps :wink:

Reply 19

Totally Tom

18

yeh sorry i just missed an x off.

durr.

integrate (lnx)^2

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