February 17th, 2018, 09:58 PM  #1 
Newbie Joined: Feb 2018 From: India Posts: 5 Thanks: 0 Math Focus: calculas, linear algebra, geometry, complex analysis,polynomials  Calculus: Integration
${\displaystyle\int_0^1}\! \sqrt{x\ln\small\left(\frac{1}{\large x}\right)}dx$
Last edited by greg1313; February 17th, 2018 at 10:56 PM. 
February 17th, 2018, 10:57 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 18,719 Thanks: 1536 
According to WA, it's $\displaystyle \frac13\sqrt{\frac{2\pi}{3}}$.

February 18th, 2018, 07:08 AM  #3 
Newbie Joined: Feb 2018 From: India Posts: 5 Thanks: 0 Math Focus: calculas, linear algebra, geometry, complex analysis,polynomials 
elaborate solution

February 18th, 2018, 08:21 AM  #4 
Newbie Joined: Feb 2018 From: India Posts: 5 Thanks: 0 Math Focus: calculas, linear algebra, geometry, complex analysis,polynomials 
also, if square root is replaced by cube root what will be the solution?

February 18th, 2018, 09:42 AM  #5 
Global Moderator Joined: Dec 2006 Posts: 18,719 Thanks: 1536 
Have you tried using a substitution?

February 18th, 2018, 09:43 AM  #6 
Senior Member Joined: Sep 2015 From: USA Posts: 1,790 Thanks: 923  
February 18th, 2018, 06:30 PM  #7 
Global Moderator Joined: Dec 2006 Posts: 18,719 Thanks: 1536 
Using $x = e^{u^2}$, where $u\geqslant0$, so that $dx = 2ue^{u^2}du$, one gets $\displaystyle \int_\infty^0\! 2u^2e^{3u^2/2}du = (2/3)\!\int_0^\infty\! 3u^2e^{3u^2/2}du$. Using integration by parts, with $v = e^{(3/2)u^2}$, so that $dv = 3ue^{(3/2)u^2}du$, one gets $\displaystyle (2/3)\left[ue^{(3/2)u^2}\right]_0^\infty + (2/3)\!\int_0^\infty\! e^{(3/2)u^2}du = (2/3)\!\int_0^\infty\! e^{(3/2)u^2}du$. If $u = \sqrt{\frac23}t$, so that $du = \sqrt{\frac23}dt$, one gets $\frac23\sqrt{\frac23}$ $\displaystyle \!\int_0^\infty\! e^{t^2}dt$. It's well known that $\displaystyle \!\int_0^\infty\! e^{t^2}dt = \small\frac{\sqrt\pi}{2}$. 