Discussion
\(\displaystyle\int e^{x}\left[\frac{1}{1 + x} - \frac{1}{(1 + x)^{2}}\right] dx =\)
Answer: A
The integral is \(\frac{e^{x}}{1 + x} + C\).
The integrand has the form \(e^{x}[f(x) + f'(x)]\) with \(f(x) = \frac{1}{1 + x}\), since \(f'(x) = -\frac{1}{(1 + x)^{2}}\). And
\[ \int e^{x}[f(x) + f'(x)]\,dx = e^{x}f(x) + C \]
because \(\frac{d}{dx}[e^{x}f(x)] = e^{x}(f + f')\). The other options do not differentiate back to the integrand; for example \(\frac{xe^{x}}{1 + x}\) has derivative \(\frac{e^{x}(x^{2} + x + 1)}{(1 + x)^{2}}\).
The pattern \(e^{x}[f(x) + f'(x)]\) is worth spotting quickly: whenever the bracket contains a function and its derivative, the integral is \(e^{x}\) times that function.
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