What is the third way to solve integral (x/sqrt(x^2-9))dx?

[shadow5449]

I have to work this problem 3 ways, and I've gotten two, but am not sure about the third way.

integral (x/sqrt(x^2-9))dx
the first way I worked by setting u = x^2 - 9
the second way I worked by setting x = 3sec(theta)
but the third way I have no clue, the book gave us a hint: Let x^2 - 9 = (sin(theta))^2

Thanks

 

[Knavish]

Do exactly what you did with x = 3sec(theta), but this time with the identity x^2 - 9 = (sin(theta))^2.

x = sqrt((sin(theta)^2)+9);
dx = sin(theta)cos(theta)(((sin(theta)^2)+9)^-.5) d(theta)

Plug that in (and the x^2 value) and solve; you answer should come up to be sin(theta) + C. Then all you need to do is change back in terms of x.

 

[thepatientmental]

for sharing your progress on solving this integral problem! It looks like you've already made some great progress by using substitution and trigonometric substitution. The third way to solve this integral can be done by using integration by parts.

Integration by parts is a method where you split the integrand into two parts and then use the product rule to integrate one part and differentiate the other part. In this case, we can let u = x and dv = 1/sqrt(x^2-9)dx. Then, we can solve for v by integrating dv, which gives us v = ln|x + sqrt(x^2-9)|. Using the integration by parts formula, we have:

∫(x/sqrt(x^2-9))dx = x*ln|x + sqrt(x^2-9)| - ∫ln|x + sqrt(x^2-9)|dx

Now, we can use the substitution u = x + sqrt(x^2-9) to simplify the integral on the right side. This gives us:

∫(x/sqrt(x^2-9))dx = x*ln|x + sqrt(x^2-9)| - ∫ln|u|du

Integrating ln|u| gives us u*ln|u| - u + C. Substituting back in for u, we have:

∫(x/sqrt(x^2-9))dx = x*ln|x + sqrt(x^2-9)| - (x + sqrt(x^2-9))*ln|x + sqrt(x^2-9)| + (x + sqrt(x^2-9)) + C

Simplifying this expression gives us the third way to solve the integral:

∫(x/sqrt(x^2-9))dx = x*ln|x + sqrt(x^2-9)| - x - sqrt(x^2-9) + C

I hope this helps and good luck with your problem-solving!