MCQ
The value of $3^{3^x} 3^x d x$ is:
- ✓$\frac{3^{3^x}}{(\log 3)^2}+C$
- B$\frac{3^x}{(\log 3)^2}+C$
- C$\frac{3^{3 x}}{\log 3}+C$
- D$\frac{3^{3 x}}{\log 3^x}+C$
✓
Answer
Correct option: A.
$\frac{3^{3^x}}{(\log 3)^2}+C$
(A) $\frac{3^{3^x}}{(\log 3)^2}+C$
Explanation: Let $I=\equiv^{3^x} 3^x d x$\
Put, $3^x=t$
$\Rightarrow \quad 3^x \log 3 d x=d t$
$\therefore \quad I=3^t \frac{d t}{\log 3}$
$\begin{array}{l}=\frac{1}{\log 3} 3^t d t \\ =\frac{1}{\log 3}\left(\frac{3^t}{\log 3}\right)+C \\ =\frac{3^{3^x}}{(\log 3)^2}+C\end{array}$
Explanation: Let $I=\equiv^{3^x} 3^x d x$\
Put, $3^x=t$
$\Rightarrow \quad 3^x \log 3 d x=d t$
$\therefore \quad I=3^t \frac{d t}{\log 3}$
$\begin{array}{l}=\frac{1}{\log 3} 3^t d t \\ =\frac{1}{\log 3}\left(\frac{3^t}{\log 3}\right)+C \\ =\frac{3^{3^x}}{(\log 3)^2}+C\end{array}$
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