Q115 / 180 · COMEDK 2025 (10 May Shift 3) · Mathematics
$\int\left(e^{x \log _e 6}\right) e^x d x=\phi(x)+c$ then $\phi(x)=$
- A$\dfrac{6^x}{1+\log _e 6}$
- B$\dfrac{e^x}{\log 6 e}$
- Correct answerC$\dfrac{(6 e)^x}{1+\log _e 6}$
- D$6^x e^x$
Solution
The integral is given by $I = \int (e^{x \log_{e} 6}) e^x dx$.
Using the property $e^{x \log_{e} a} = (e^{\log_{e} a})^x = a^x$, we have $e^{x \log_{e} 6} = 6^x$.
Substituting this into the integral, $I = \int 6^x e^x dx = \int (6e)^x dx$.
Using the standard integral formula $\int a^x dx = \dfrac{a^x}{\log_{e} a} + c$, where $a = 6e$, we get:
$I = \dfrac{(6e)^x}{\log_{e} (6e)} + c$.
Using the property $\log_{e} (ab) = \log_{e} a + \log_{e} b$, we have $\log_{e} (6e) = \log_{e} 6 + \log_{e} e = \log_{e} 6 + 1$.
Therefore, $\phi(x) = \dfrac{(6e)^x}{1 + \log_{e} 6}$.
Answer: $\dfrac{(6 e)^x}{1+\log _e 6}$
Using the property $e^{x \log_{e} a} = (e^{\log_{e} a})^x = a^x$, we have $e^{x \log_{e} 6} = 6^x$.
Substituting this into the integral, $I = \int 6^x e^x dx = \int (6e)^x dx$.
Using the standard integral formula $\int a^x dx = \dfrac{a^x}{\log_{e} a} + c$, where $a = 6e$, we get:
$I = \dfrac{(6e)^x}{\log_{e} (6e)} + c$.
Using the property $\log_{e} (ab) = \log_{e} a + \log_{e} b$, we have $\log_{e} (6e) = \log_{e} 6 + \log_{e} e = \log_{e} 6 + 1$.
Therefore, $\phi(x) = \dfrac{(6e)^x}{1 + \log_{e} 6}$.
Answer: $\dfrac{(6 e)^x}{1+\log _e 6}$
Hence, the correct answer is option: 'C'.