Can anyone help me in integrating this simple equation ???????????????????????????????????????????????

Question

(log x) / (1+x ^ 0.333)


You should take x ^ 0.333 as cube root of x. Also explain each and every step included in integration (solution of this problem).



But remember that you are not allowed to write directly the answer, you should solve it and prove yourself right.



If you solve it directly then your answer will be declared as invalid
and if you give full details including step - by - step detailed solution then you will be given full 10 points i.e., your answer will be chosen as the best answer.


Do you understand?


So, be as detailed as possible. More the detailed you are, more the points you will get.


Remember, to get full 10 points you have to write at least 12 lines in your solution and you should explain each and every step.

In fact this question is a mean to test your knowledge about integration, a useful branch of physics, chemistry and mathematics (some portion is used in biology also). Be quick and detailed to get full points

Answer 1

So sorry, this integral is not elementary. It's
solution involves the dilogarithm function.
Let's break it down and see what happens.
First, let u = x^(1/3)
x = u³, dx = 3u² du
Then the integral becomes
3∫ ln u³/(1+ u) * u² du
which simplifies to
9∫ ln u *(u²/(1+u) ) du
Now use long division to write
u²/(1+u) = u - u/(u+1)
So we finally have to integrate
9*∫ u ln u du - 9*∫ ln u * u/(u+1) du.
Again u/(u+1) = 1 - 1/(u+1).
by long division.
So our task is now to compute
9*[ ∫u ln u du - ∫ ln u du - ∫ ln u/(u+1) du ].
Let's do them one at a time. We can do the first 2
by parts, but the last one is the one involving the
dilogarithm function.
A). ∫ u ln u du
Let U = ln u dV = u du
then
dU = 1/u V = u²/2
So A). yields u²/2 ln u - ∫ u/2 du = u²/2 ln u - u²/4

B). ∫ ln u du
Use parts again
U = ln u dV = du
dU = du/u v = u
So B gives us
u ln u - ∫ du = u(ln u -1)

C). ∫ ln u/(u1) dx
Again let's try to use integration by parts.
Let U = ln u dV = 1/(u+1) du
Then
dU = 1/u v = ln(u+1)
So this becomes
ln u * ln(u+1) - ∫ ln(u+1)/u du.
Finally, writing t = -u, du = -dt,
we see that this last integral is just the integral
of -dilog(-u). (See definition on Wikipedia website.)
Hope that helps a bit!

Answer 2

Sorr y cant be very detailed rite now.. but hope u understand this i got the same ans as d other person:
use the dilogarithm function.
Let's break it down and see what happens.
let u = x^(1/3)
x = u³, dx = 3u² du
integral:
3∫ ln u³/(1+ u) * u² du
=9∫ ln u *(u²/(1+u) ) du
write by l.div
u²/(1+u) = u - u/(u+1)
integrate
9*∫ u ln u du - 9*∫ ln u * u/(u+1) du.
one more time
=u/(u+1) = 1 - 1/(u+1).
by L. division.
done
9*[ ∫u ln u du - ∫ ln u du - ∫ ln u/(u+1) du ].
one by one
first 2 by parts,
last one is use d dilogarithm function.
A). ∫ u ln u du
Let U = ln u dV = u du
this means
dU = 1/u V = u²/2
therefore A). yields u²/2 ln u - ∫ u/2 du = u²/2 ln u - u²/4

B). ∫ ln u du
remember
U = ln u dV = du
dU = du/u v = u
So B=
u ln u - ∫ du = u(ln u -1)

C). ∫ ln u/(u1) dx
Again let's try to use integration by parts.
let U = ln u dV = 1/(u+1) du
dU = 1/u v = ln(u+1)
this is
ln u * ln(u+1) - ∫ ln(u+1)/u du.
now t = -u, du = -dt,
done

Answer 3

f(x) = ln(x)*x^-1/3
Integrating by parts Int = 3/2x^2/3*Ln(x) -3/2Int x^2/3/x=

3/2*x^2/3[Lnx-3/2]

Answer 4

forget integration

Answer 5

too busy to answer the question

Answer 6

mathematics professor can answer you

Answer 7

3x=3*x are you mad