COMEDK · Maths · 28. Indefinite Integration
\(\int \dfrac{3^x}{\sqrt{1-9^x}} d x\) is equal to
- A \((\log 3) \sin ^{-1} 3^x+C\)
- B \(\dfrac{1}{3} \sin ^{-1}\left(3^x\right)+C\)
- C \(\dfrac{1}{\log 3} \sin ^{-1} 3^x+C\)
- D \(3 \log 3 \sin ^{-1} 3^x+C\)
Answer & Solution
Correct Answer
(C) \(\dfrac{1}{\log 3} \sin ^{-1} 3^x+C\)
Step-by-step Solution
Detailed explanation
Let \(I = \int \dfrac{3^x}{\sqrt{1-9^x}} dx\). Rewrite the denominator as \(\sqrt{1-(3^x)^2}\). Let \(u = 3^x\). Then \(du = 3^x \ln 3 dx\), which implies \(3^x dx = \dfrac{du}{\ln 3}\). Substituting these into the integral:…
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