Question

How would you integrate `int_1^e 1/(x sqrt(ln^2x))dx` ?

`int_1^e 1/(x sqrt(ln^2x))dx`

Answer 1

This integral does not exist.

Explanation:

Since `ln x>0` in the interval `[1,e]`, we have

`sqrt{ln^2 x}= |ln x| = ln x`

here, so that the integral becomes

`int_1^e dx/{x ln x}`

Substitute `ln x = u`, then `dx/x = du` so that

`int_1^e dx/{x ln x} = int_{ln 1}^{ln e} {du}/u = int_0^1 {du}/u`

This is an improper integral, since the integrand diverges at the lower limit. This is defined as

`lim_{l -> 0^+} int_l^1 {du}/u`

if this exists. Now

`int_l ^1 {du}/u = ln 1 - ln l = -ln l`

since this diverges in the limit `l -> 0^+`, the integral does not exist.

Answer 2

`pi/2`

Explanation:

The integral `int_1^e("d"x)/(xsqrt(1-ln^2(x))`.

Substitute first `u=ln(x)` and `"d"u=("d"x)/x`.

Thus, we have
`int_(x=1)^(x=e)("d"u)/sqrt(1-u^2)`

Now, substitute `u=sin(v)` and `"d"u=cos(v)"d"v`.

Then,
`int_(x=1)^(x=e)(cos(v))/(sqrt(1-sin^2(v)))\ "d"v=int_(x=1)^(x=e)"d"v` since `1-sin^2(v)=cos^2(v)`.

Continuing, we have
`[v]_(x=1)^(x=e)=[arcsin(u)]_(x=1)^(x=e)=[arcsin(ln(x))]_(x=1)^(x=e)=arcsin(ln(e))-arcsin(ln(1))=pi/2-0=pi/2`