Question

$\int \frac{2^x+4^x-2x\cdot 7^x}{7^x}dx$ is equal to
(where $C$ is integration constant)

• A. $\frac{{\left(\frac{2}{7}\right)}^x}{\ln \left(\frac{2}{7}\right)}+\frac{{\left(\frac{4}{7}\right)}^x}{\ln \left(\frac{4}{7}\right)}-x^2+C$
• B. $\frac{{\left(\frac{2}{7}\right)}^{-x}}{\ln \left(\frac{2}{7}\right)}+\frac{{\left(\frac{4}{7}\right)}^{-x}}{\ln \left(\frac{4}{7}\right)}+x^2+C$
• C. $\frac{{\left(\frac{2}{7}\right)}^x}{\ln \left(\frac{2}{7}\right)}-\frac{{\left(\frac{4}{7}\right)}^x}{\ln \left(\frac{4}{7}\right)}+x^2+C$
• D. $-\frac{{\left(\frac{2}{7}\right)}^x}{\ln \left(\frac{2}{7}\right)}-\frac{{\left(\frac{4}{7}\right)}^x}{\ln \left(\frac{4}{7}\right)}-x^2+C$

Answer 1

The correct option is A $\frac{{\left(\frac{2}{7}\right)}^x}{\ln \left(\frac{2}{7}\right)}+\frac{{\left(\frac{4}{7}\right)}^x}{\ln \left(\frac{4}{7}\right)}-x^2+C$

$\int \frac{2^x+4^x-2x\cdot 7^x}{7^x}dx$
$=\int {\left(\frac{2}{7}\right)}^{x}dx+\int {\left(\frac{4}{7}\right)}^x-2\int xdx\{\because \int a^{x}dx=\frac{a^x}{\ln a}+C\}=\frac{{\left(\frac{2}{7}\right)}^x}{\ln \left(\frac{2}{7}\right)}+\frac{{\left(\frac{4}{7}\right)}^x}{\ln \left(\frac{4}{7}\right)}-x^2+C$