Answer:
[tex]\frac{x^2-16}{2} + 16ln\frac{4}{x} +16C[/tex]
Step-by-step explanation:
Given the indefinite integral [tex]\int\limits{\frac{x^2-16}{x} } \, dx[/tex], using the substitute
x = 4 sec(θ)...1
The integral can be calculated as thus;
First let us diffrentiate the substitute function with respect to θ
dx/dθ = 4secθtanθ
dx = 4secθtanθdθ... 2
Substituting equation 1 and 2 into the integral function we will have;
[tex]\int\limits{\frac{(4sec \theta)^2-16}{4sec \theta} } \, 4sec \theta tan \theta d \theta\\\int\limits{\frac{16sec^2 \theta-16}{4sec \theta} } \, 4sec \theta tan \theta d \theta\\\int\limits{\frac{(16(sec^2 \theta-1)}{4sec \theta} } \, 4sec \theta tan \theta d \theta\\\\from \ trig \ identity,\ sec^2 \theta - 1 = tan^\theta\\\\\int\limits{\frac{16 tan^2 \theta}{4sec \theta} } \, 4sec \theta tan \theta d \theta\\\\\int\limits 16 tan^3 \theta d \theta\\\\[/tex]
Find the remaining solution in the attachment.
