please help integral of xlnxdx from 1 to + infinite..step by step?

Answer 1

Use integration by parts.

First, see the functions as the product of two functions:

(x) * (lnx)

Define one as "u" and the other as "dv".
and take the derivative of "u" and the integral of "dv"

u = lnx du = 1/x dx

dv = x v = (x^2)/2


The formula uses the above information:

Integral(x * lnx * dx) = uv - [Integral (vdu)]

=lnx * (x^2)/2 - [Integral ((x^2)/2 * 1/x dx)]
=lnx * (x^2)/2 - [Integral (x/2 dx)]
=lnx * (x^2)/2 - (x^2)/4
From 1 to infinity = infinity

Answer 2

To integrate it, you have to use integration by parts.
Let u = lnx and dv = x dx
du = (1/x) dx and v = (1/2)x^2
udv = uv - vdu
= (1/2)x^2 lnx - integral [(1/2)(x^2)(1/x) dx]
= (1/2)x^2 lnx - integral [(1/2)x dx]
= (1/2)x^2 lnx - (1/4)x^2
Replace infinity with b and take the limit as b goes to infinity.
= limit b->infinity[ [(1/2)b^2 lnb - (1/4)b^2] - [(1/2)1^2 ln(1) -
(1/4)1^2]]
= limit b->infinity [[(1/2)b^2 lnb - (1/4)b^2] - 1/4]
= limit b->infinity [b^2[(1/2)lnb - (1/4)] - 1/4]
This does not have a limit, so the integral does not exist.

Answer 3

Clearly this integral is infinite since xlnx goes to infinity as x does.

Answer 4

Do a u substitution. x + a million = u x = u - a million dx = du replace into the equation. [(u-a million)^2 + 2]/u du = (u^2 - 2u + a million + 2)/u du = u^2/u - 2u/u +3/u du = u - 2 + 3/u du now combine u^2/2 - 2u + 3*ln(u) + C replace x back in (x+a million)^2/2 - 2(x+a million) + 3ln(x+a million) + C uncertain in the journey that your instructor needs you to simplify yet i'm hoping this facilitates!

Answer 5

this integral is infinite.

for x≥1 xlnx≥0

for x≥e xlnx≥1

thus ∫(from 1 to ∞)xlnxdx ≥ ∫(from e to ∞)dx + ∫(from 1 to e)xlnxdx ≥ ∫(from e to ∞)dx =∞

Answer 6

Absurd, infinity