ex (f(x) + f1 (x) dx) =
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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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