How to find #int((1)/(x-sqrt(x+2)))dx# ?

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Answer 1 · 2018-03-09T08:34:20

The answer is #=2/3ln(sqrt(x+2)+1)+4/3ln(|sqrt(x+2)-2|)+C#

Make the substitution

#u=sqrt(x+2)#, #=>#, #du=1/(2sqrt(x+2))*dx#

#x=u^2-2#

Therefore,

#int(dx)/(x-sqrt(x+2))=int(2udu)/(u^2-2-u)#

#=2int(udu)/(u^2-u-2)#

#u^2-u-2=(u+1)(u-2)#

Perform the decomposition into partial fractions

#u/((u+1)(u-2))=A/(u+1)+B/(u-2)#

#=(A(u-2)+B(u+1))/((u+1)(u-2))#

The denominators are the same, compare the numerators

#u=A(u-2)+B(u+1)#

Let #u=-1#, #=>#, #-1=-3A#, #=>#, #A=1/3#

Let #u=2#, #=>#, #2=3B#, #=>#, #B=2/3#

Therefore,

#u/((u+1)(u-2))=(1/3)/(u+1)+(2/3)/(u-2)#

So,

#2int(udu)/(u^2-u-2)=2int((1/3)/(u+1)+(2/3)/(u-2))du#

#=2/3ln(u+1)+4/3ln(u-2)#

#=2/3ln(sqrt(x+2)+1)+4/3ln(|sqrt(x+2)-2|)+C#