How you calculate this? #int_0^1sqrtx/((x+3)sqrt(x+3))#, Using substitution of #sqrt(x/(x+3))=t#.

1 Answer
mason m
May 13, 2017

#-1+ln3#

Explanation:

Establishing a couple identities first from #sqrt(x/(x+3))=t#:

#t^2=x/(x+3)" "" "=>" "" "(1/(x+3)=t^2/x" "color(blue)((star)))#

#1/t^2=(x+3)/x=1+3/x#

#3/x=1/t^2-1=(1-t^2)/t^2#

#x/3=t^2/(1-t^2)#

#x=(3t^2)/(1-t^2)" "color(red)((star))#

#dx=(6t(1-t^2)-3t^2(-2t))/(1-t^2)^2dt=(6t)/(1-t^2)^2dt" "color(green)((star))#

Combining #color(blue)((star))# with #color(red)((star))#:

#1/(x+3)=t^2/((3t^2)/(1-t^2))=(1-t^2)/3" "color(orange)((star))#

Then we have the integral:

#intsqrtx/((x+3)sqrt(x+3))dx=int1/(x+3)sqrt(x/(x+3))dx#

#=int(1-t^2)/3(t)(6t)/(1-t^2)^2dt=int(6t^2)/(3(1-t^2))dt#

#=2intt^2/(1-t^2)dt=2int(t^2-1+1)/(1-t^2)#

#=2int(-(1-t^2))/(1-t^2)dt-2int1/(t^2-1)dt#

#=-2intdt-2int1/((t+1)(t-1))dt#

Partial fraction decomposition on the latter integral gives:

#=-2t+int1/(t+1)dt-int1/(t-1)dt#

#=-2t+lnabs(t+1)-lnabs(t-1)#

#=-2sqrt(x/(x+3))+lnabs( ( sqrt(x/(x+3))+1) / ( sqrt(x/(x+3))-1) )#

#=-2sqrt(x/(x+3))+lnabs((sqrtx+sqrt(x+3))/(sqrtx-sqrt(x+3)))#

Now applying the bounds:

#int_0^1sqrtx/((x+3)sqrt(x+3))dx=(-2sqrt(x/(x+3))+lnabs((sqrtx+sqrt(x+3))/(sqrtx-sqrt(x+3))))|_0^1#

#=(-2sqrt(1/4)+lnabs((1+sqrt4)/(1-sqrt4)))-(2(0)+lnabs((0+sqrt3)/(0-sqrt3)))#

#=-2(1/2)+lnabs(-3)-(0+ln1)#

#=-1+ln3#

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