If ∫2x+1−5x−110xdx ==A15xlog⁡15+B12xlog⁡12+c then

I=∫2x+1−5x−110xdx=∫5x 2x25x−15· 2x10xdx=∫215x−1512xdx=215xlog⁡15−1512xlog⁡12+c formula ∫axdx=axloge⁡a+c

If ∫2x+1−5x−110xdx ==A15xlog⁡15+B12xlog⁡12+c then

I=∫2x+1−5x−110xdx=∫5x 2x25x−15· 2x10xdx=∫215x−1512xdx=215xlog⁡15−1512xlog⁡12+c formula ∫axdx=axloge⁡a+c

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$If \int \frac{2^{x + 1} - 5^{x - 1}}{10^{x}} d x = = A \frac{{(\frac{1}{5})}^{x}}{\log (\frac{1}{5})} + B \frac{{(\frac{1}{2})}^{x}}{\log (\frac{1}{2})} + c t h e n$

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$A = \frac{2}{5}, B = \frac{1}{5}$

$A = \frac{2}{5}, B = \frac{- 1}{5}$

$A = 2, B = - \frac{1}{5}$

$A = 2, B = \frac{1}{5}$

answer is C.

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Detailed Solution

$\begin{array}{l} I = \int \frac{2^{x + 1} - 5^{x - 1}}{10^{x}} d x \\ = \int \frac{5^{x} 2^{x} (\frac{2}{5^{x}} - \frac{1}{5 \cdot 2^{x}})}{10^{x}} d x \\ = \int [2 {(\frac{1}{5})}^{x} - \frac{1}{5} {(\frac{1}{2})}^{x}] d x \\ = \frac{2 {(\frac{1}{5})}^{x}}{\log (\frac{1}{5})} - \frac{1}{5} \frac{{(\frac{1}{2})}^{x}}{\log (\frac{1}{2})} + c f o r m u l a \int a^{x} d x = \frac{a^{x}}{\log_{e} a} + c \end{array}$