How do I integrate #intdx/(sqrt(x)(sqrt(x)+1))#?

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Answer 1 · 2017-10-29T03:27:12

#int dx/[sqrtx*(sqrtx+1)]=2Ln(sqrtx+1)+C#

#I=int dx/[sqrtx*(sqrtx+1)]#

After using #x=(tanu)^4# and #dx=4(tanu)^3*(secu)^2*du# transforms, #I# became,

#I=int (4(tanu)^3*(secu)^2*du)/[(tanu)^2*(secu)^2]#

=#int 4tanu*du#

=#4ln(secu)+C#

=#2Ln[(secu)^2]+C#

=#2Ln[(tanu)^2+1]+C#

After using #x=(tanu)^4# and #(tanu)^2=sqrt(x)# inverse transforms, I found,

#I=int dx/[sqrtx*(sqrtx+1)]#

=#2Ln(sqrtx+1)+C#

Note: This integral also can be found by #sqrt(x)=u#, #x=u^2# and #dx=2u*du# transforms,

Answer 2 · 2017-10-29T03:35:41

#2ln(sqrtx+1)+C#

Use the substitutions:

#u = sqrtx + 1#

#du = dx/(2sqrtx)#

#intdx/(sqrtx(sqrtx+1)) = 2intdx/(2sqrtx(sqrtx+1)) = 2int(du)/u = 2ln|u|+C#

#= 2ln(sqrtx+1)+C#

Final Answer