Q82 / 180 · COMEDK 2023 (Evening Shift) · Mathematics
$\int x^x(1+\log x) d x$ is equal to
- Correct answerA$x^x+c$
- B$x \log x+c$
- C$x^{x-1}+c$
- D$x^x \log x+c$
Solution
Let $I = \int x^x(1 + \log x) dx$.
Consider the function $f(x) = x^x$.
To differentiate $f(x)$, take the natural logarithm on both sides: $\ln(f(x)) = \ln(x^x) = x \ln x$.
Differentiating both sides with respect to $x$: $\dfrac{1}{f(x)} \cdot f'(x) = \dfrac{d}{dx}(x \ln x) = x \cdot \dfrac{1}{x} + \ln x \cdot 1 = 1 + \log x$.
Therefore, $f'(x) = f(x)(1 + \log x) = x^x(1 + \log x)$.
Since the derivative of $x^x$ is $x^x(1 + \log x)$, the integral of $x^x(1 + \log x)$ is $x^x + c$.
Answer: $x^x+c$
Consider the function $f(x) = x^x$.
To differentiate $f(x)$, take the natural logarithm on both sides: $\ln(f(x)) = \ln(x^x) = x \ln x$.
Differentiating both sides with respect to $x$: $\dfrac{1}{f(x)} \cdot f'(x) = \dfrac{d}{dx}(x \ln x) = x \cdot \dfrac{1}{x} + \ln x \cdot 1 = 1 + \log x$.
Therefore, $f'(x) = f(x)(1 + \log x) = x^x(1 + \log x)$.
Since the derivative of $x^x$ is $x^x(1 + \log x)$, the integral of $x^x(1 + \log x)$ is $x^x + c$.
Answer: $x^x+c$
Hence, the correct answer is option: 'A'.