Question

How do you evaluate the definite integral `int dx / ( x(sqrt(ln(x)))` from `[e, e^81]`?

Answer 1

`int_e^(e^81)dx/(xsqrtln(x))=16`

Explanation:

We want to find:

`int_e^(e^81)dx/(xsqrtln(x))`

We will want to use substitution. Note that if we let `u=ln(x)`, then `(du)/dx=1/x` and `du=dx/x`, which is currently present in the integral.

Before making these substitutions, note that making the substitutions will necessitate that the boundaries change. To change them, plug the current bounds into `u=ln(x)`.

We see that `e` becomes `u(e)=ln(e)=1` and `e^81` becomes `u(e^81)=ln(e^81)=81`.

Thus, since `color(red)(u=ln(x))` and `color(blue)(du=dx/x)`:

`int_e^(e^81)color(blue)dx/(color(blue)xsqrtcolor(red)ln(x))=int_1^81(du)/sqrtu`

Rewrite the integral:

`int_1^81(du)/sqrtu=int_1^81u^(-1/2)du`

Which can be integrated using the rule:

`int_a^bu^ndu=[u^(n+1)/(n+1)]_a^b" "" ",n!=-1`

So:

`int_1^81u^(-1/2)du=[u^(-1/2+1)/(-1/2+1)]_1^81=[2sqrtu]_1^81=2sqrt81-2sqrt1=16`

Answer 2

16

Explanation:

just as an alternative approach,

for `int_e^(e^81)dx/(xsqrtln(x))`

if you recognise the pattern

`d/dx (sqrtln(x) )`

`= 1/2 1/ sqrt (ln( x)) * 1/x`

then the integral is

`int_e^(e^81) d/dx (2 sqrtln(x)) \ dx`

`=2 [ sqrtln(x) ]_e^(e^81)`

`2( [ sqrt(81)] - [ sqrt(1) ] ) = 16`