How do you follow this through? Integral as x goes to 1 to infinity of 1/x dx?

Answer 1

inf
∫(1/x)dx
1

= ln(x) with limits of 1 (lower limit) and infinity (upper limit)
= ln(infinity) - ln(1)
= infinity - 0
= infinity

Answer 2

Integral as x goes to 1 to infinity of 1/x dx
=lnx with limits 1 to infinity=infinity-0=infinity answer

Answer 3

Integral (1 to infinity, 1/x dx )

Since this is an improper integral (as they are referring to infinity), your first step is to express this as a limit.

lim Integral (1 to t, 1/x dx )
t -> infinity

Now, solve the integral. 1/x is a known derivative; it is the derivative of ln(x). Therefore, our integral is

lim (ln|x|) {from 1 to t}
t -> infinity

And we evaluate the function at the bounds.

lim (ln|t| - ln|1|)
t -> infinity

lim (ln|t| - 0)
t -> infinity

lim ln|t|
t -> infinity

As t approaches infinity, ln|t| approaches infinity.
Therefore, the answer is infinity.

***
Improper integrals must always be expressed as limits. Seeing a bound of negative infinity or infinity is one indication of an improper integral. Here is another example of another, less obvious improper integral:

Integral (-3 to 3, (1/x^2) dx )

This is an improper integral because our bounds go from -3 to 3, and we have a discontinuity at x = 0. We must split this integral in 2 pieces; each being one-sided limits.

lim (-3 to t, 1/x^2 dx) + lim (t to 3, 1/x^2 dx)
t -> 0- . . . . . . . . . . . . . . .t -> 0+

Answer 4

Integral as x goes to 1 to infinity of 1/x dx
=limit as s-->infinity of Integral as x goes to 1 to s of 1/x dx
=limit as s-->infinity of ln(s)
=infinity

Answer 5

Integrating 1/x gets you Ln(x)

However the limit of Ln(x) as x tends to infinity is infinite, so the integral on this interval is non-convergent.

It is possible to have integrals over infinite intervals, but there must be a limit of the integral as x tend to positive or negative infinity. Then you work it through like a normal problem and subtract the lower limit from the upper limit.

Answer 6

Is infinite (integral of 1/x dx = Lnx) so the value = ln(infinity) - ln(1) = Ln(infinity) = infinity.

Answer 7

integ from 1 to infinity (dx/x)
=ln x evaluated from 1 to infinity
= ln infinity - ln1 = infinity

Answer 8

at the beginning, you probably did no longer take the crucial wisely. ? (a million+x) / (x²+a million) dx (dx / x² + a million + ?x / x² + a million arctan(x) +(ln(x²+a million)) / 2 could be your crucial as quickly as you plug interior the bounds from 0 to infinity, you do see that the crucial diverges. to uncertain a thank you to instruct the Cauchy theory nevertheless.

Answer 9

omfg your a nerd