How to integrate 1/x^2sqrtx^2+25 dx ?

1 Answer
Apr 23, 2018

#I=-sqrt(x^2+25)/(25x)+c#

Explanation:

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YOUR QUESTION between two hash sign:

#1/x^2sqrtx^2+25 dx#.

ANSWER for :

#I=int1/(x^2sqrt(x^2+25))dx#

Let,

#x=5tanu=>dx=5sec^2udu andtanu=x/5#

So,

#I=int(5sec^2u)/(25tan^2usqrt(25tan^2u+25))du#

#=1/25int(5sec^2u)/(tan^2u5sqrt(tan^2u+1))du#

#=1/25int(5sec^2u)/(tan^2u(5secu))du#

#=1/25intsecu/tan^2udu#

#=1/25int(1/cosu)/(sin^2u/cos^2u)du#

#=1/25intcosu/sin^2udu#

#=1/25int(cscucotu)du#

#I=1/25(-cscu)+c...to(A)#

Now, #cscu=1/sinu=1/(tanucosu)=secu/tanu=sqrt(tan^2u+1)/tanu#

#i.e.cscu=sqrt((x^2/25)+1)/(x/5)=sqrt(x^2+25)/x#

Hence, from #(A)#

#I=1/25(-sqrt(x^2+25)/x)+c#

#=-sqrt(x^2+25)/(25x)+c#