Integration using substitution #intsqrt(1+x^2)/x dx#?How do I solve this question, please help me?

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Answer 1 · 2018-03-22T09:05:52

#sqrt(1+x^2)-1/2ln(abs(sqrt(1+x^2)+1))+1/2ln(abs(sqrt(1+x^2)-1))+C#

Use #u^2=1+x^2#, #x=sqrt(u^2-1)#
#2u(du)/(dx)=2x#, #dx=(udu)/x#

#intsqrt(1+x^2)/xdx=int(usqrt(1+x^2))/x^2du#

#intu^2/(u^2-1)du=int1+1/(u^2-1)du#

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

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

#u=1#
#1=2B#, #B=1/2#

#u=-1#
#1=-2A#, #A=-1/2#

#int1-1/(2(u+1))+1/(2(u-1))du=u-1/2ln(abs(u+1))+1/2ln(abs(u-1))+C#

Putting #u=sqrt(1+x^2)# back in gives:
#sqrt(1+x^2)-1/2ln(abs(sqrt(1+x^2)+1))+1/2ln(abs(sqrt(1+x^2)-1))+C#