MHT CET · Maths · Indefinite Integration
\(\quad \int 3^{3^x} \cdot 3^x d x=\)
- A \(\frac{3^x}{(\log 3)^2}+\mathrm{c}\), where c is a constant of integration.
- B \(\frac{3^{3^x}}{\log 3}+\mathrm{c}\), where c is a constant of integration.
- C \(\frac{3^{3^x}}{(\log 3)^2}+\mathrm{c}\), where c is a constant of integration.
- D \(\frac{3^x}{\log 3}+\mathrm{c}\), where c is a constant of integration.
Answer & Solution
Correct Answer
(C) \(\frac{3^{3^x}}{(\log 3)^2}+\mathrm{c}\), where c is a constant of integration.
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \text { Let } \mathrm{I}=\int 3^{3^x} \cdot 3^x \mathrm{~d} x \\ & \text { Let } 3^x=\mathrm{t} \\ & 3^x \log 3 \mathrm{~d} x=\mathrm{dt} \\ & 3^x \mathrm{~d} x=\frac{1}{\log 3} \mathrm{dt} \\ & \therefore \quad \mathrm{I}=\int 3^{\mathrm{t}} \cdot \frac{1}{\log 3} \cdot \mathrm{dt} \\ & \quad=\frac{1}{\log 3} \int 3^{\mathrm{t}} \mathrm{dt} \\ & \quad=\frac{1}{\log 3} \times \frac{3^{\mathrm{t}}}{\log 3}+\mathrm{c} \\ & \mathrm{I}=\frac{3^{3^x}}{(\log 3)^2}+\mathrm{c}\end{aligned}\)
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