Question

How do you evaluate the definite integral `int (dx/(xsqrtlnx))` from `[1,e^3]`?

Answer 1

`2sqrt3`

Explanation:

We have the definite integral:

`int_1^(e^3)dx/(xsqrtlnx)`

This is a prime time to use substitution. Note that in the integrand, we have both `lnx` and its derivative, `1/x`. An added bonus is that letting `u=lnx` makes dealing with the square root a lot easier.

So, if we let `u=lnx`, then `du=dx/x`. We can substitute these both into the integrand easily, and we will in a second, but not before we change the bounds of the definite integral.

Since we are moving from an integral in terms of `x` (since we have `dx`) to an integral in terms of `u` (thanks to `du`), we have to change the bounds of the integral. To do so, plug the current bounds into `u=lnx`.

Thus the bound of `1` becomes `u(1)=ln(1)=0`, and the bound of `3` becomes `u(3)=ln(e^3)=3`.

This integral is about to change dramatically. We see that:

`int_1^(e^3)dx/(xsqrtlnx)=int_0^3(du)/sqrtu`

In order to integrate this, rewrite `1/sqrtu` with fractional exponents. Then, use the integration rule `intu^ndu=u^(n+1)/(n+1)+C`, where `n!=-1`.

`int_0^3(du)/sqrtu=int_0^3u^(-1/2)du=[u^(-1/2+1)/(-1/2+1)]_0^3=[u^(1/2)/(1/2)]_0^3=[2sqrtu]_0^3`

`=2sqrt3-2sqrt0=2sqrt3`