How to use tabular method to integrate ∫ x Inx dx?

Question

I know how to integrate and differentiate using tabular but it seems i'm stuck on how to integrate Inx?
Please don't tell me to use integration by parts to integrate Inx or use integration by parts to solve ∫ x Inx dx? I want to know what is the integral of Inx to carry on with the tabular format. Thx

Answer 1

The antiderivative of ln x is x*(ln x - 1), which you derive by plain integration by parts. (u = ln x, dv = dx).



x ... ln x

1 ... x(ln x - 1)

0 ... [∫ x Inx dx] - x^2/2



So ∫ x Inx dx = x^2 (ln x - 1) + [∫ x Inx dx] - x^2/2



Sometimes when you loop back to the original, you get a different sign or leading coefficient, leaving a final equation that you can "solve" for the integral. Not so here. The tabular method doesn't work for this integral.



In finding out what the tabular method was (I don't recall seeing in when I took Calculus), I found this web page.



http://math.ucsd.edu/~wgarner/math20b/int_by_parts.htm



It does a pretty good job of explaining the method, and introduces an acronym LIPET for choosing the u in u dv. Tend to prefer (L)ogarithms, (I)nverse functions, (P)olynomials, (E)xponential functions, or (T)rigonometric functions,...in that order. The tabular method only works when u is the polynomial, so that would tend to steer you away from that method when logs or inverse functions (like arctan x, arcsinh x, etc) are involved.

Answer 2

Tabular Integration

Answer 3

You didn't want integral by parts, but I'll do the work for ∫ln x dx so you're convinced about the solution:



∫ ln x dx---> integral by parts, gives, if u = ln x, dv = dx; then,

du = 1/x dx, v =x

∫ ln x dx = uv - ∫v du

= x ln x - ∫ (x) (1/x) dx

= x ln x - ∫ 1 dx

= x ln x - x or x ( ln x - 1 ).



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Backcheck:

d ( x ln x - x) = {x (d [ ln x]) + (ln x) dx} - 1 dx

= x ( 1/x) dx + ln x dx - dx

= dx + ln x dx - dx

= ln x.

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Is that what your question was? Because you clearly asked "I want to know what is the integral of ln x"... so that's what it is, ∫ ln x = x ln x - x.