Evaluate the integral 16t3 5t2 271 5t2 1 t2 2 dxSolutio

Evaluate the integral.

16t3 - 5t2 + 271 - 5/(t2 + 1) (t2 + 2) dx

Solution

Use the quotient rule: If f(x)=g(x)/h(x) then the derivative of f(x) is: f\'(x)=[g\'(x)h(x)-g(x)h\'(x)]/[h(x)^2] In this case, g(x) = 16t^3 - 5t^2 + 27t - 5 and h(x) = (t^2 + 1) (t^2 + 2) = t^4 + 3t^2 + 2 g\'(x) = 48t^2 - 10t +27 h\'(x) = 4t^3 + 6t Now plug these values into the quotient rule to get: [(48t^2 - 10t +27)(t^4 + 3t^2 + 2) - (16t^3 - 5t^2 + 27t)(4t^3 + 6t)]/[(4t^3 + 6t)^2] as the derivative! You can simplify as necessary. Hope this helps! :) Let me know if you have any more questions!