Question

$\int \frac{2^{x+1}-5^{x-1}}{{10}^x}dx$ is equal to
(where $C$ is constant of integration)

• A. $\frac{2{\left(\frac{1}{5}\right)}^x}{\ln \left(\frac{1}{5}\right)}-\frac{1}{5}\frac{{\left(\frac{1}{2}\right)}^x}{\ln \left(\frac{1}{2}\right)}+C$
• B. $\frac{3{\left(\frac{1}{5}\right)}^x}{\ln \left(\frac{1}{5}\right)}-\frac{1}{5}\frac{{\left(\frac{1}{2}\right)}^x}{\ln \left(\frac{1}{2}\right)}+C$
• C. $\frac{4{\left(\frac{1}{5}\right)}^x}{\ln \left(\frac{1}{5}\right)}-\frac{1}{5}\frac{{\left(\frac{1}{2}\right)}^x}{\ln \left(\frac{1}{2}\right)}+C$
• D. $\frac{5{\left(\frac{1}{5}\right)}^x}{\ln \left(\frac{1}{5}\right)}-\frac{1}{5}\frac{{\left(\frac{1}{2}\right)}^x}{\ln \left(\frac{1}{2}\right)}+C$

Answer 1

The correct option is A $\frac{2{\left(\frac{1}{5}\right)}^x}{\ln \left(\frac{1}{5}\right)}-\frac{1}{5}\frac{{\left(\frac{1}{2}\right)}^x}{\ln \left(\frac{1}{2}\right)}+C$

$I=\int \frac{2^{x+1}-5^{x-1}}{{10}^x}dx=\int [2{\left(\frac{1}{5}\right)}^x-\frac{1}{5}{\left(\frac{1}{2}\right)}^x]dx$
$\{\because \int a^x=\frac{a^x}{\ln a}\}$
$\Rightarrow I=\frac{2{\left(\frac{1}{5}\right)}^x}{\ln \left(\frac{1}{5}\right)}-\frac{1}{5}\frac{{\left(\frac{1}{2}\right)}^x}{\ln \left(\frac{1}{2}\right)}+C$