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√ ex(f(x) + f'(x) ) dx = |
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Answer» Answer: To BEGIN with, let’s say 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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