How do you find the indefinite integral of #int 1/(1+sqrt(2x))#?

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Answer 1 · 2016-11-19T06:18:57

The answer is #(1+sqrt(2x))-ln(1+sqrt(2x))+C#

We do this integral by substitution

Let #u=1+sqrt(2)sqrtx#

#sqrtx=(u-1)/sqrt2#

Then, #du=sqrt2*dx/(2sqrtx)=dx/(sqrt2sqrtx)#

#dx=sqrt2du*sqrtx=sqrt2*du*(u-1)/sqrt2#

#dx=(u-1)du#

Therefore,

#intdx/(1+sqrt()2x)=int((u-1)du)/u#

#=int(1-1/u)du=u-lnu#

#intdx/(1+sqrt()2x)=(1+sqrt(2x))-ln(1+sqrt(2x))+C#