How do you evaluate the integral #int 1/(xsqrt(2+x^2))#?
1 Answer
Explanation:
#I=intdx/(xsqrt(2+x^2))#
Use the substitution
Plugging this in shows that:
#I=int(sqrt2sec^2thetad theta)/(sqrt2tanthetasqrt(2+2tan^2theta))#
Note that
#I=int(sqrt2sec^2thetad theta)/(sqrt2tantheta(sqrt2sectheta))=1/sqrt2intsectheta/tanthetad theta#
Rewriting with
#I=1/sqrt2int1/costhetacostheta/sinthetad theta=1/sqrt2intcscthetad theta=-1/sqrt2lnabs(csctheta+cottheta)#
From the original substitution we have
Thus,
So:
#I=-1/sqrt2lnabs((sqrt(x^2+2)+sqrt2)/x)+C#
We can rewrite this using
#I=1/sqrt2lnabs(x/(sqrt(x^2+2)+sqrt2))+C#