TS EAMCET · Maths · Indefinite Integration
If \(\frac{3 \pi}{4} < x < \frac{7 \pi}{4}\) then \(\int\left(2^x-\sqrt{1+\sin 2 x}+\frac{1}{x^2}-\frac{1}{x}\right) d x=\)
- A \(\frac{2^x}{\log 2}-\sin x+\cos x-\frac{1}{x}-\log x+c\)
- B \(2^x \log 2+\sin x-\cos x-\frac{1}{x}+\frac{1}{x^2}+c\)
- C \(\frac{2^x}{\log 2}+\sin x-\cos x-\frac{1}{x}-\log x+c\)
- D \(2 \log 2-\sin x+\cos x-\frac{1}{+}+\frac{1}{+c}\)
Answer & Solution
Correct Answer
(C) \(\frac{2^x}{\log 2}+\sin x-\cos x-\frac{1}{x}-\log x+c\)
Step-by-step Solution
Detailed explanation
Given integral is \(\int\left(2^x-\sqrt{1+\sin 2 x+\frac{1}{x^2}-\frac{1}{x}}\right) d x\). \(I=\int\left(2^x-\sqrt{|\sin x+\cos x|^2}+\frac{1}{x^2}-\frac{1}{x}\right) \cdot d x\)…
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