MCQ
Evaluate: $\int\left(2^x+2^{-x}\right)^2 d x$
  • A
    $\frac{1}{2 \log 2}\left(2^{2 x}-2^{-2 x}\right)+C$
  • $\frac{1}{2 \log 2}\left(2^{2 x}-2^{-2 x}\right)+2 x+C$
  • C
    $\frac{1}{2 \log 2}\left(2^{2 x}+2^{-2 x}\right)+2 x+C$
  • D
    $\frac{1}{2 \log 2}\left(2^{2 x}+2^{-2 x}\right)+C$

Answer

Correct option: B.
$\frac{1}{2 \log 2}\left(2^{2 x}-2^{-2 x}\right)+2 x+C$
We have$, \int\left(2^x+2^{-x}\right)^2 d x=\int\left(2^{2 x}+2^{-2 x}+2\right) d x$
$=\frac{2^{2 x}}{(\log 2) \times 2}+\frac{2^{-2 x}}{(\log 2)(-2)}+2 \cdot x+C$
$=\frac{1}{2 \log 2}\left(2^{2 x}-2^{-2 x}\right)+2 x+C$

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