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

Question

Emilia Aguilar · Accepted Answer

#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#

Charles Arocha · Answer

undefined To find the indefinite integral of $ \int \frac{\sqrt{169x^2 - 81}}{x} \, dx $, follow these steps:

1. Recognize that the integrand can be simplified using trigonometric substitution.

2. Let $ x = \frac{9}{13} \sec(\theta) $.

3. Find $ dx $ using the derivative of $ \sec(\theta) $, which is $ \sec(\theta) \tan(\theta) \, d\theta $.

4. Substitute $ x $ and $ dx $ in terms of $ \theta $ into the integral.

5. Simplify the integrand using trigonometric identities.

6. Integrate with respect to $ \theta $.

7. Substitute back $ x $ in terms of $ \theta $ to obtain the result in terms of $ x $.