How do you find the indefinite integral of #int sqrt(169x^2–81)/x dx#?

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Answer 1 · 2018-05-01T17:09:26

#I=sqrt(169x^2-81)-9sec^-1((13x)/9)+c#

Here,

#I=intsqrt(169x^2-81)/x dx#

#=9intsqrt((169/81)x^2-1)/x dx#

Let,

#13/9x=secu=>13/9dx=secutanudu#

# =>dx=9/13secutanudu and u=sec^-1((13x)/9)#

So,

#I=9intsqrt(sec^2u-1)/(9/13secu)xx9/13secutanudu#

#=9inttanu/secuxxsecutanudu#

#=9inttan^2udu#

#=9int(sec^2u-1)du#

#=9(tanu-u)+c#

#=9(sqrt(sec^2u-1)-u)+c#

Subst. back ,#secu=(13x)/9 and u=sec^-1((13x)/9)#

#I=9[sqrt((169/81x^2-1))-sec^-1((13x)/9)]+c#

#=9[(sqrt(169x^2-81)/9-sec^-1((13x)/9)]+c#

#I=sqrt(169x^2-81)-9sec^-1((13x)/9)+c#