# Find the integral of ln(x)/sqrt(x) as x is between 0 and 1

Hi Im trying to figure out the following..Ive tried substitutions but cant find any that work:

$$\int_0^1 \frac{lnx}{ \sqrt x}dx$$

HA @ 0 so
$$\int_0^1 \frac{lnx}{ \sqrt x} dx$$
=
$$\lim_{t\rightarrow 0^+} \int_t^1 \frac{lnx}{\sqrt x}dx$$

Last edited:

StatusX
Homework Helper
Try the substitution u=sqrt(x).

so....2du=1/root(x)dx.. it becomes just ln(u^2)?

if this is right I end up with

$$\lim_{t\rightarrow 0^+} 2 - \frac{s\sqrt t}{t}$$

so would this be divergent?

StatusX
Homework Helper
I don't know what s is there, but yes, this is divergent. Another way to find this is to note that this function is less than ln(x), whose integral diverges to negative infinity on this interval.

ok, the s should be a two...so basically if the integral ever has $$\frac{a}{0}$$...then it is called divergent?..or whenever an answer cannot be found..it is divergent?

StatusX
Homework Helper
This integral is divergent because the limit you mention in your first post is infinity. "if the integral ever has a/0" isn't precise enough an expression for me to say anything meaningful about. In general, for improper integrals like this one, you have to take a limit as you approach vertical asymptopes or as [itex]x \rightarrow \pm \infty[/tex]. If the result after taking all these limits is infinity, the integral is said to diverge. Whether you can find the answer or not is your problem, not the integral's.