COMEDK · Maths · 28. Indefinite Integration
\(\int \dfrac{2^{x}}{\sqrt{1-4^{x}}} d x\) is equal to
- A \((\log 2) \sin ^{-1} 2^{x}+C\)
- B \(\dfrac{1}{2} \sin ^{-1} 2^{x}+C\)
- C \(\dfrac{1}{\log 2} \sin ^{-1} 2^{x}+C\)
- D \(2 \log 2 \sin ^{-1} 2^{x}+C\)
Answer & Solution
Correct Answer
(C) \(\dfrac{1}{\log 2} \sin ^{-1} 2^{x}+C\)
Step-by-step Solution
Detailed explanation
Let \(I = \int \dfrac{2^{x}}{\sqrt{1-4^{x}}} dx\). Rewrite the integral as \(I = \int \dfrac{2^{x}}{\sqrt{1-(2^{x})^{2}}} dx\). Substitute \(u = 2^{x}\). Then \(du = 2^{x} \ln 2 \, dx\), which implies \(2^{x} dx = \dfrac{du}{\ln 2}\). Substituting these into the integral gives…
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