
William W. answered 11/28/23
Experienced Tutor and Retired Engineer
You wouldn't typically use a u-sub for the first problem - in my perspective it makes it harder.
∫(7ex + 5)/ex dx = ∫7ex/ex dx + ∫5/ex dx = ∫7 dx + ∫5e-x dx = 7x - 5e-x + C
In the second example, let u = ex, then du/dx = ex or rewriting it, dx = du/ex or dx = du/u
So the integral becomes:
∫u/(7u + 5) du/u = ∫1/(7u+5) du = (1/7)ln(7u + 5) + C = (1/7)ln(7ex + 5)