1. ## [SOLVED] Improper Integral

Last one!!

$\displaystyle \int\limits^{1}_{0} 35 \frac{ln(x)}{\sqrt x} dx$

Please don't give me the entire problem, I just can't figure out what technique I need to use to get started. Should I just do a substitution? I tried doing u=lnx, but that square root is throwing me.
Thanks!

2. Originally Posted by mollymcf2009
Last one!!

$\displaystyle \int\limits^{1}_{0} 35 \frac{ln(x)}{\sqrt x} dx$

Please don't give me the entire problem, I just can't figure out what technique I need to use to get started. Should I just do a substitution? I tried doing u=lnx, but that square root is throwing me.
Thanks!
Your pick for the substitution is correct.

Let $\displaystyle u=\ln x\implies \,du=\frac{1}{x}\,dx$

The trick here is to note that $\displaystyle 35\int_0^1\frac{\ln x}{\sqrt{x}}\,dx=35\int_0^1\frac{\sqrt{x}\ln x}{x}\,dx$

So now when you apply the substitution, the integral becomes $\displaystyle 35\int_0^1 u\sqrt{x}\,du$

Now to get rid of $\displaystyle \sqrt{x}$, recall that we let $\displaystyle u=\ln x\implies e^u=x\implies \sqrt{x}=e^{\frac{1}{2}u}$

Thus, you integral (including change of limits) becomes $\displaystyle 35\int_{-\infty}^0 ue^{\frac{1}{2}u}\,du$

Can you take it from here?