\(\displaystyle\int \frac{e^{x}}{(e^{x} + 1)(e^{x} + 2)}\,dx\) is equal to (where C is an arbitrary constant)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\log\left(\frac{e^{x} + 1}{e^{x} + 2}\right) + C\)
The numerator \(e^{x}\,dx\) is the differential of \(e^{x}\), so we substitute
\(t = e^{x}\), \(dt = e^{x}\,dx\)
The integral becomes
\( \displaystyle \int \frac{dt}{(t + 1)(t + 2)} \)
By partial fractions:
\( \displaystyle \frac{1}{(t + 1)(t + 2)} = \frac{1}{t + 1} - \frac{1}{t + 2} \)
So
\( \displaystyle \int \frac{dt}{(t + 1)(t + 2)} = \log(t + 1) - \log(t + 2) + C \)
\( \displaystyle = \log\left(\frac{t + 1}{t + 2}\right) + C \)
(No modulus is needed, since t = ex > 0.) Putting back \(t = e^{x}\):
\( \displaystyle \int \frac{e^{x}}{(e^{x} + 1)(e^{x} + 2)}\,dx = \log\left(\frac{e^{x} + 1}{e^{x} + 2}\right) + C \)