- 2 x^3
We find the derivative of an integral using the Second Fundamental Theorem of Calculus:
f(t) dt = f(x). Normally, we would need to rewrite the absolute value function as a
piecewise function, but notice that we are evaluating the absolute value over an interval where
all values will be positive. Thus, we can ignore the absolute value and rewrite the integral as
|t| dt = t dt. We get t dt = (x^2 )(2x) = 2x^3 . Don’t forget that because the
upper limit is a function, we need to multiply the answer by the derivative of that function.
- sin (ln x) + C
Recall that ln x = . Here, we can use u-substitution to get rid of the log in the integrand.
If we let u = ln x, then du = dx. If we substitute into the integrand, we get ∫ cos (ln x) dx =
∫ cos u du = sin u + C. Substituting back, we get ∫^ cos (ln x) dx = sin (ln x) + C.
- sin (2 + ex) + C
Recall that ∫eu du = eu + C. Let’s use u-substitution. If we let u = 2 + ex, then du = ex dx. If we
substitute into the integrand, we get ∫ex cos (2 + ex) dx = ∫cos u du = sin u + C. Substituting
back, we get ∫ex cos (2 + ex) dx = sin (2 + ex) + C.
tan−1(e^3 x) + C
Recall that ∫ = tan−1 u + C. Here we have ∫ , and we will need to do u-
