COMEDK · Maths · 28. Indefinite Integration
\(\int \frac{4^x}{\sqrt{1-16^x}} d x\) is equal to
- A \((\log 4) \sin ^{-1} 4^x+C\)
- B \(\frac{1}{4} \sin ^{-1}\left(4^x\right)+C\)
- C \(\frac{1}{\log 4} \sin ^{-1} 4^x+C\)
- D \(4 \log 4 \sin ^{-1} 4+C\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{\log 4} \sin ^{-1} 4^x+C\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} I & =\int \frac{4^x}{\sqrt{1-16^x}} d x \\ & =\int \frac{4^x}{\sqrt{1-\left(4^x\right)^2}} d x \end{aligned}\) On putting, \(4^x=t\), we get \(4^x \log 4 d x=d t\)…
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