Evaluate the integral with hyperbolic or trigonometric substitution. ?

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1 Answer
Mar 29, 2018

#-1/4ln(csc(tan^-1(x/4))+cot(tan^-1(x/4)))+C#

Explanation:

by using the substitution :
#x=4tanu#
#dx=4sec^2u*du#

#1+tan^2u=sec^2u#

#intdx/(x*sqrt(16+x^2))#=#int(4sec^2u*du)/(4tanu*4secu)#

#=1/4int(secu*du)/tanu#
#secu=1/cosu#

#tanu=sinu/cosu#

by simplifying:
=#1/4intcscu*du#

#=-1/4ln(cscu+cotu)+C#
by substituting again
#u=tan^-1(x/4)#

you get :
#intdx/(x*sqrt(16+x^2))#=#-1/4ln(csc(tan^-1(x/4))+cot(tan^-1(x/4)))+C#