What is the integral of x.lnx?

Answer 1

Do you want ∫x lnx dx?

Let's call that I, which we now seek to find.

The solution is I = ½ x2 lnx - ¼ x2;

here is how we can find it:

Let z = lnx.

Then, x = ez,

dx = ez dz, and

dI = x lnx dx.

Then, dI = (ez)(z)(ez dz)

= ze2z dz

= ¼ (2z e2z d(2z)).

Thus, 4dI = wew dw,

where we let w = 2z = 2 lnx.

Now, d(wew) = ew dw + w d(ew)

= ew dw + wew dw

= ew dw + 4dI; hence,

4dI = d(wew) - ewdw

= d(wew) - d(ew)­ = d((w - 1)ew).

Therefore, 4I = (w - 1)ew

= (2 lnx - 1)x2

= 2x2 lnx - x2; and

I = ½ x2 lnx - ¼ x2,

which is the answer we sought.

Checking, we differentiate back,

to confirm that I' = x lnx:

I = ¼ x2(2 lnx - 1), whence,

4I' = (x2(2 lnx - 1))'

= 2x(2 lnx - 1) + x2 (2/x)

= 2x(2 lnx - 1) + 2x

= 2x(2 lnx)

= 4x lnx; thus,

I' = x lnx,

re-assuring us that we have integrated correctly.