Question

$\int e^{-x}(1+(1-x)\log x)dx$ is equal to

• A. $(x\log x)e^{x}+C$
• B. $(x^{2}\log x)e^{-x}+C$
• C. $(x^{2}\log x)e^{-2x}+C$
• D. $(x\log x)e^{-x}+C$
• E. $(x\log x)e^{-2x}+C$

Hint:

Logic Tip: In competitive exams, if the integrand looks like a complicated mix of exponentials and algebraic/logarithmic functions, immediately check the options. Differentiating a product rule expression takes seconds, whereas setting up multiple layers of Integration by Parts can take minutes.

Answer 1

The Correct Option is D

Concept:
For complex integrals, especially those in multiple-choice exams involving $e^x$ or $e^{-x}$ multiplied by other functions, evaluating by differentiating the given options is often much faster and less error-prone than applying Integration by Parts. Integration is the reverse process of differentiation.
Step 1: Analyze the structure of the integrand.
The given integrand is $e^{-x}(1 + (1-x)\log x)$. Expanding this gives: $e^{-x} + e^{-x}\log x - x e^{-x}\log x$. Notice the recurring components $x$, $\log x$, and $e^{-x}$. This suggests the original function before differentiation was a product of these terms.
Step 2: Test Option D by differentiation.
Let's differentiate the expression in Option D: $y = (x\log x)e^{-x}$. We apply the Product Rule $(uv)' = u'v + uv'$, where $u = x\log x$ and $v = e^{-x}$. First, find $u'$ using the Product Rule again: $u' = \frac{d}{dx}(x\log x) = (1)(\log x) + (x)\left(\frac{1}{x}\right) = \log x + 1$ Second, find $v'$: $v' = \frac{d}{dx}(e^{-x}) = -e^{-x}$
Step 3: Combine the derivative terms.
$$y' = (\log x + 1)(e^{-x}) + (x\log x)(-e^{-x})$$
Step 4: Factor out $e^{-x}$ to match the integrand.
$$y' = e^{-x} [ (\log x + 1) - x\log x ]$$ Group the $\log x$ terms together: $$y' = e^{-x} [ 1 + \log x (1 - x) ]$$ This exactly matches the integrand provided in the question. Thus, Option D is the correct antiderivative.