1.

Ex (f(x) + f1 (x) dx) =​

Answer»

Step-by-step EXPLANATION:

Integral of the TYPE e^x[F(x) + f ‘(x)]dx

To begin with, let’s say

I = ∫ ex [f(x) + f ’(x)] dx

Opening the brackets, we get,

I = ∫ ex f(x) dx + ∫ ex f ’(x) dx = I1 + ∫ ex f ’(x) dx … (1)

Where, I1 = ∫ ex f(x) dx

To solve I1, we will use integration by parts. Let the FIRST function = f1(x) = f(x) and the second function = g1(x) = ex. Therefore,

I1 = f(x) ∫ ex dx – ∫ [df(x)/dx ∫ ex dx] dx

Or, I1 = ex f(x) – ∫ ex f ’(x) dx + C

Substituting the value of I1 in equation (1), we get

I = ex f(x) – ∫ ex f ’(x) dx + ∫ ex f ’(x) dx + C = ex f(x) + C

Thus, ∫ ex [f(x) + f ’(x)] dx = ex f(x) + C … (2)


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