`int(2^(x)3^(x))/(1+(36)^(x))dx=`

Options

A. `(tan^(-1)(6^(x)))/(log6)+c`
B. `(1)/(12)log((1+6^(x))/(1-6^(x)))+c`
C. `(1)/(12)log((6^(x)-1)/(6^(x)+1))+c`
D. `(1)/(12)tan^(-1)((6^(x))/(log6))+c`

Answer

To solve the integral

\[ I = \int \frac{2^x 3^x}{1 + 36^x} \, dx, \]

we can follow these steps:

### Step 1: Simplify the integrand
We start by rewriting the expression in the integrand. Notice that:

\[ 2^x 3^x = (2 \cdot 3)^x = 6^x, \]
and

\[ 36^x = (6^2)^x = 6^{2x}. \]

Thus, we can rewrite the integral as:

\[ I = \int \frac{6^x}{1 + 6^{2x}} \, dx. \]

### Step 2: Use substitution
Next, we will use the substitution \( t = 6^x \). Then, we need to find \( dx \) in terms of \( dt \).

Differentiating \( t = 6^x \) gives:

\[ dt = 6^x \ln(6) \, dx \quad \Rightarrow \quad dx = \frac{dt}{6^x \ln(6)} = \frac{dt}{t \ln(6)}. \]

### Step 3: Substitute in the integral
Now we substitute \( t \) and \( dx \) into the integral:

\[ I = \int \frac{t}{1 + t^2} \cdot \frac{dt}{t \ln(6)} = \frac{1}{\ln(6)} \int \frac{1}{1 + t^2} \, dt. \]

### Step 4: Integrate
The integral

\[ \int \frac{1}{1 + t^2} \, dt \]

is a standard integral that evaluates to

\[ \tan^{-1}(t) + C. \]

Thus, we have:

\[ I = \frac{1}{\ln(6)} \left( \tan^{-1}(t) + C \right). \]

### Step 5: Substitute back for \( t \)
Now we substitute back \( t = 6^x \):

\[ I = \frac{1}{\ln(6)} \left( \tan^{-1}(6^x) + C \right). \]

### Final Result
Therefore, the final result of the integral is:

\[ I = \frac{1}{\ln(6)} \tan^{-1}(6^x) + C. \]

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