First slide

Introduction to integration

Question

$\int \frac{2^{x + 1} - 5^{x - 1}}{10^{x}} d x$ equals

Easy

A

$\frac{1}{5 \log 2} 2^{x + 1} - \frac{2}{\log 5} 5^{x - 1} + C$
B

$\frac{2 (x + 1)}{5^{x + 1}} - \frac{x}{2^{x}} + C$
C

$\frac{2 (x + 1)}{5^{x + 1}} + \frac{x}{2^{x}} + C$
D

$\frac{1}{5 \log 2} 2^{- x} - \frac{2}{\log 5} 5^{- x} + C$

Solution

$\begin{aligned} \int \frac{2^{x + 1} - 5^{x - 1}}{10^{x}} d x = \int ({(\frac{2}{10})}^{x} \cdot 2 - {(\frac{5}{10})}^{x} \frac{1}{5}) d x \\ = 2 \int {(\frac{1}{5})}^{x} d x - \frac{1}{5} \int {(\frac{1}{2})}^{x} d x \\ = 2 \frac{5^{- x}}{- \log 5} + \frac{1}{5} \frac{2^{- x}}{\log 2} + C \end{aligned}$

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