Question

The value of ${\int }_0^4{3}^{\sqrt{2x+1}}dx$ is

• A. $\frac{6}{\log 3}(13-\frac{4}{\log 3})$
• B. $\frac{66}{\log 3}$
• C. $\frac{6}{\log 3}(13-\frac{5}{\log 3})$
• D. $\frac{78}{\log 3}+\frac{24}{(\log 3{)}^2}$

Answer 1

The correct option is A $\frac{6}{\log 3}(13-\frac{4}{\log 3})$

${\int }_0^4{3}^{\sqrt{2x+1}}dx=\int a^{x}dx=\frac{a^x}{\log a}={\int }_0^4{3}^{\sqrt{2x+1}}dx=[\frac{3^{\sqrt{2x+1}}}{\log 3}\times \frac{1}{\frac{d}{dx}\sqrt{2x+1}}{]}_0^4=[\frac{3^{\sqrt{2x+1}}}{\log 3}\times 2\sqrt{2x+1}\times a{]}_0^4=\frac{6}{\log 3}(13-\frac{4}{\log 3})$