TS EAMCET · Maths · Indefinite Integration
\(\int \frac{3^x d x}{\sqrt{9^x-1}}\) is equal to
- A \(\frac{1}{\log 3} \log \left|3^x+\sqrt{9^x-1}\right|+c\)
- B \(\frac{1}{\log 3} \log \left|3^x-\sqrt{9^x-1}\right|+c\)
- C \(\frac{1}{\log 9} \log \left|3^x+\sqrt{9^x-1}\right|+c\)
- D \(\frac{1}{\log 3} \log \left|9^x+\sqrt{9^x-1}\right|+c\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{\log 3} \log \left|3^x+\sqrt{9^x-1}\right|+c\)
Step-by-step Solution
Detailed explanation
Let \(I=\int \frac{3^x d x}{\sqrt{9^x-1}}=\int \frac{3^x}{\sqrt{3^{2 x}-1}} d x\) Let \(3^x=z \Rightarrow 3^x \log 3 d x=d z\) \[ I=\frac{1}{\log 3} \int \frac{z}{\sqrt{z^2-1}} \]…
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