Integrate (lnx)^3? some one must know the technique.?

Answer 1

DL Pilgrim has the right idea. Let's simplify it a bit.
Doing this by parts over and over is soo tedious!
Let's find a simpler way.
Let u = ln x
x = e^u
dx = e^u du.
So our integral is now
∫ u³ e^u du.
Since the 4th derivative of u³ is 0, we can use tic-tac-toe here:
u³ e^u
3u² e^u
6u e^u
6 e^u
0 e^u
So ∫ u³ e^u du = e^u(u³ -3u² + 6u -6) + C
Now substitute back:
∫ (ln x)³ dx = x( (ln x)³ -3*(ln x)² + 6 ln x -6) + C

Answer 2

Let u=lnx
so du=1/x dx
and dx = xdu = e^u du

Then integral (lnx)^3 dx
= integral u^3 * e^u du

Now integrate by part using formula
integral adb= ab - integral bda
with u^3 = a and e^u = b

So integral is u^3 e^u - integral 3u^2 e^u

You now have to integrate 3u^2 e^u again by parts (and it will have to be done yet again a third time after this), and you will eventually clear away the products in the integration and get down to something that you can fully integrate. Don't forget to add C at the end!

I hope this helps a a guideline of what to do.

Answer 3

Integral (ln x)^3 dx

x = e^u
dx = e^u du

Integral u^3 * e^u du

e^u du = dv
u^3 = w
e^u = v
3u^2 du = dw

(e^u)(u^3) - 3 Integral (u^2)(e^u) du

Integral (u^2)(e^u) du

e^u du = dv
u^2 = w
e^u = v
2u du = dw

Integral (u^2)(e^u) du = (e^u)(u^2) - 2 Integral u e^u du

Integral u e^u

e^u du = dv
u = w
e^u = v
du = dw

(e^u)(u) - Integral e^u du) = (e^u)(u) - e^u + C
-----
Answer: (e^u)(u^3) - 3(e^u)(u^2) + 6(e^u)(u) - 6e^u + C

Check some of the sign changes for the answer because I had to substitute within a subtitution of a substition....

Answer 4

you want: int ln(x) ^ 3 dx, right?
try this:

u = ln(x) so e^u = x
du = dx/x, so

dx = x du or
= e^u du

the integral is then

int u^3 e^u du, try int by parts

w = u^3
dw = 3 u^2 du

dv = e^u du
v = e^u

wv - int v dw

u^3 e^u - int e^u * 3 u^2 du. Long, but will maybe work.

so we have

u^3 e^u - 3 int e^u * u^2 du, and we need to work on this last part. Do int by parts again:

Answer 5

was off the day we did integration, but...
by inspection gen form looks recursive
ie INT(lnx)^n=x*(lnx)^n-n*INT((lnx)^n-1))dx
So all you need to do do is successively subst n=3,2,1 in formula above
eg INT(lnx)^2=x*ln2^2-2*INT(lnx)dx etc
(INT= integr8 sign btw)

Answer 6

Repeated parts formula:

first stage: let u = (ln x)³ and dv/dx = 1

That involves an integral involving (ln x)²

same trick again, letting dv/dx = 1

once more and you'll get rid of the ln x

Don't forget the + c



Solution:

∫ (ln x)³dx = x(ln x)³ - ∫x3(ln x)²(1/x)dx

= x(ln x)³ - 3∫(ln x)²dx

= x(ln x)³ - 3{x(ln x)² - ∫x2(ln x)(1/x)dx}

= x(ln x)³ - 3x(ln x)² + 6∫(ln x)dx

= x(ln x)³ - 3x(ln x)² + 6{x(ln x) - ∫x(1/x)dx}

= x(ln x)³ - 3x(ln x)² + 6x(ln x) - 6∫dx

= x(ln x)³ - 3x(ln x)² + 6x(ln x) - 6x + c

No need for any substitution.