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

1 Answer
Mar 9, 2018

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

Explanation:

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