Question

${\int }_0^1\frac{3^{x+1}-4^{x-1}}{{12}^x}dx=$

• A. $\frac{9}{4}lo{g}_{4}e$
• B. $\frac{9}{4}lo{g}_{4}e-\frac{1}{6}lo{g}_{3}e$
• C. $lo{g}_{4}3$
• D. None of these

Answer 1

The correct option is B $\frac{9}{4}lo{g}_{4}e-\frac{1}{6}lo{g}_{3}e$

${\int }_0^1\frac{3^{x+1}-4^{x-1}}{{12}^x}dx$ $={\int }_0^1(\frac{3}{4^x}-\frac{1}{4}.\frac{1}{3^x})dx$
$={\{3\frac{\frac{1}{4^x}}{\left(log\frac{1}{4}\right)}-\frac{1}{4}\frac{\frac{1}{3^x}}{log\left(\frac{1}{3}\right)}\}}_0^1$
$=\frac{3(\frac{1}{4}-1)}{-{\log }_{e}4}-\frac{1}{4}\frac{(\frac{1}{3}-1)}{-{\log }_{e}3}$
$=\frac{9}{4}{\log }_{4}e-\frac{1}{6}{\log }_{3}e$
Ans: B