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# How do you find the integral of ##int (lnx / sqrt(x)) dx## from 0 to 1?

I found: ##-4##

Let us try by changing ##1/sqrt(x)==x^(-1/2)## and then Integrate By Parts: ##int(ln(x)/sqrt(x))dx==int(x^(-1/2)ln(x))dx=## ##=2x^(1/2)ln(x)-int(2x^(1/2)*1/xdx)=## ##=2x^(1/2)ln(x)-int(2x^(1/2-1)dx)=## ##=2x^(1/2)ln(x)-int(2x^(-1/2)dx)=## ##=2x^(1/2)ln(x)-4x^(1/2)=## let us use our extrema: ##=2x^(1/2)ln(x)-4x^(1/2)|_0^1####=-4##

Only problematic part is ##2x^(1/2)lnx## when ##x->0_+##, so we will find the limit:

##lim_(x->0_+)2x^(1/2)lnx= 0*(-oo)## which is undefined.

Using the rule of L'Hospital:

##lim_(x->0_+)2x^(1/2)lnx=2lim_(x->0_+)lnx/x^(-1/2)=##

##=2lim_(x->0_+)(1/x)/(-1/2x^(-3/2))=-4lim_(x->0_+)x^(3/2)/x=##

##=-4lim_(x->0_+)sqrtx=-4*0=0##

So, the above results holds.