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If $\frac{3 \pi}{4} < x < \frac{7 \pi}{4}$ then $\int\left(2^x-\sqrt{1+\sin 2 x}+\frac{1}{x^2}-\frac{1}{x}\right) d x=$
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Verified Answer
The correct answer is:
$\frac{2^x}{\log 2}+\sin x-\cos x-\frac{1}{x}-\log x+c$
Given integral is $\int\left(2^x-\sqrt{1+\sin 2 x+\frac{1}{x^2}-\frac{1}{x}}\right) d x$.
$I=\int\left(2^x-\sqrt{|\sin x+\cos x|^2}+\frac{1}{x^2}-\frac{1}{x}\right) \cdot d x$
$$
\begin{aligned}
I & =\int\left(2^x+\sqrt{2} \sin \left(\frac{\pi}{4}+x\right)+\frac{1}{x^2}-\frac{1}{x}\right) \cdot d x \\
I & =\int 2^x d x+\sqrt{2} \int \sin \left(\frac{\pi}{4}+x\right) d x+\int \frac{d x}{x^2}-\int \frac{d x}{x} \\
& =\frac{2^x}{\log 2}+\sqrt{2} \cos \left(\frac{\pi}{4}+x\right)-\frac{1}{x}-\log +c \\
& =\frac{2^x}{\log _2}-\sqrt{2}\left(\cos \left(\frac{\pi}{4}\right) \cos x-\sin \left(\frac{\pi}{4}\right) \sin x\right)-\frac{1}{x}-\log d c+c \\
& =\frac{2^x}{\log 2}-\cos x+\sin x-\frac{1}{x}-\log x+c
\end{aligned}
$$
So, option (c) is correct.
$I=\int\left(2^x-\sqrt{|\sin x+\cos x|^2}+\frac{1}{x^2}-\frac{1}{x}\right) \cdot d x$
$$
\begin{aligned}
I & =\int\left(2^x+\sqrt{2} \sin \left(\frac{\pi}{4}+x\right)+\frac{1}{x^2}-\frac{1}{x}\right) \cdot d x \\
I & =\int 2^x d x+\sqrt{2} \int \sin \left(\frac{\pi}{4}+x\right) d x+\int \frac{d x}{x^2}-\int \frac{d x}{x} \\
& =\frac{2^x}{\log 2}+\sqrt{2} \cos \left(\frac{\pi}{4}+x\right)-\frac{1}{x}-\log +c \\
& =\frac{2^x}{\log _2}-\sqrt{2}\left(\cos \left(\frac{\pi}{4}\right) \cos x-\sin \left(\frac{\pi}{4}\right) \sin x\right)-\frac{1}{x}-\log d c+c \\
& =\frac{2^x}{\log 2}-\cos x+\sin x-\frac{1}{x}-\log x+c
\end{aligned}
$$
So, option (c) is correct.
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