`int(3^(x+1)-7^(x-1))/(21^(x))dx=K_(1)3^(-x)+K_(2)7^(-x)+C`

Options

A. `K_(1)=(1)/(7log3),K_(2)=(1)/(3log7)`
B. `K_(1)=(1)/(log3),K_(2)=-(1)/(3log7)`
C. `K_(1)=(1)/(7log3),K_(2)=-(3)/(log7)`
D. `K_(1)=(3)/(log7),K_(2)=(-7)/(log7)`

Answer

To solve the integral

\[ \int \frac{3^{(x+1)} - 7^{(x-1)}}{21^x} \, dx, \]

we can start by rewriting the denominator \(21^x\) in terms of its prime factors:

\[ 21^x = (3 \cdot 7)^x = 3^x \cdot 7^x. \]

Now, we can rewrite the integral as:

\[ \int \frac{3^{(x+1)} - 7^{(x-1)}}{3^x \cdot 7^x} \, dx. \]

Next, we can split the integral into two separate terms:

\[ \int \left( \frac{3^{(x+1)}}{3^x \cdot 7^x} - \frac{7^{(x-1)}}{3^x \cdot 7^x} \right) \, dx. \]

This simplifies to:

\[ \int \left( \frac{3^{x+1}}{3^x} \cdot \frac{1}{7^x} - \frac{7^{(x-1)}}{7^x} \cdot \frac{1}{3^x} \right) \, dx = \int \left( 3 \cdot 7^{-x} - \frac{7^{-1}}{3^x} \right) \, dx. \]

Now we can rewrite this as:

\[ \int \left( 3 \cdot 7^{-x} - \frac{1}{7} \cdot 3^{-x} \right) \, dx. \]

Next, we can integrate each term separately:

1. For the first term \(3 \cdot 7^{-x}\):

\[ \int 3 \cdot 7^{-x} \, dx = 3 \cdot \int 7^{-x} \, dx = 3 \cdot \left(-\frac{7^{-x}}{\ln 7}\right) = -\frac{3 \cdot 7^{-x}}{\ln 7}. \]

2. For the second term \(-\frac{1}{7} \cdot 3^{-x}\):

\[ \int -\frac{1}{7} \cdot 3^{-x} \, dx = -\frac{1}{7} \cdot \left(-\frac{3^{-x}}{\ln 3}\right) = \frac{3^{-x}}{7 \ln 3}. \]

Putting it all together, we have:

\[ -\frac{3 \cdot 7^{-x}}{\ln 7} + \frac{3^{-x}}{7 \ln 3} + C. \]

Now, we can express the final result in the form:

\[ K_1 \cdot 3^{-x} + K_2 \cdot 7^{-x} + C, \]

where

\[ K_1 = \frac{1}{7 \ln 3} \quad \text{and} \quad K_2 = -\frac{3}{\ln 7}. \]

Thus, the final answer is:

\[ \int \frac{3^{(x+1)} - 7^{(x-1)}}{21^x} \, dx = -\frac{3 \cdot 7^{-x}}{\ln 7} + \frac{3^{-x}}{7 \ln 3} + C. \]