Question

How to calculate this? `d/dx(int_0^(x^3)ln(t^2+1)dt)`

Answer 1

`= 3 x^2 ln(x^6+1)`

Explanation:

By the Liebnitz Integral Rule:

`d/(dx) (\int _{a(x)}^{b(x)}f(t )\ dt )=f(b(x)) * b'(x) -f(a(x)) * a'(x)`

`implies d/dx(int_0^(x^3)ln(t^2+1)dt) = 3 x^2 ln(x^6+1)`