`inttan^(-1)alphatanxdalpha` equals

Options

A. `tan^(-1)alpha,secx+c`
B. `tan^(-1)alpha.log(secx)+c`
C. `tanx{alphatan^(-1)alpha-(1)/(2)log(1+alpha^(2))}+c`
D. `tanalpha(xtan^(-1)x-(1)/(2)log(1+x^(2)))+c`

Answer

To solve the integral \( I = \int \tan^{-1}(\alpha) \tan(x) \, d\alpha \), we can use integration by parts. Let's go through the steps systematically.

### Step 1: Identify the Parts
We will use the integration by parts formula:
\[ \int u \, dv = uv - \int v \, du \]
Here, we can let:
- \( u = \tan^{-1}(\alpha) \) (which we will differentiate)
- \( dv = \tan(x) \, d\alpha \) (which we will integrate)

### Step 2: Differentiate \( u \) and Integrate \( dv \)
Now we differentiate \( u \) and integrate \( dv \):
- \( du = \frac{1}{1 + \alpha^2} \, d\alpha \)
- \( v = \tan(x) \cdot \alpha \)

### Step 3: Apply Integration by Parts
Now we apply the integration by parts formula:
\[ I = \tan^{-1}(\alpha) \cdot \tan(x) \cdot \alpha - \int \tan(x) \cdot \alpha \cdot \frac{1}{1 + \alpha^2} \, d\alpha \]

### Step 4: Simplify the Integral
The remaining integral can be simplified:
\[ \int \tan(x) \cdot \frac{\alpha}{1 + \alpha^2} \, d\alpha \]
To solve this integral, we can use a substitution. Let:
- \( t = 1 + \alpha^2 \)
- Then, \( dt = 2\alpha \, d\alpha \) or \( d\alpha = \frac{dt}{2\alpha} \)

### Step 5: Change of Variables
Substituting \( \alpha = \sqrt{t - 1} \):
\[ \int \tan(x) \cdot \frac{\sqrt{t - 1}}{t} \cdot \frac{dt}{2\sqrt{t - 1}} = \frac{1}{2} \int \frac{\tan(x)}{t} \, dt \]
This integral is:
\[ \frac{1}{2} \tan(x) \ln|t| + C = \frac{1}{2} \tan(x) \ln|1 + \alpha^2| + C \]

### Step 6: Combine Results
Now we substitute back into our expression for \( I \):
\[ I = \tan^{-1}(\alpha) \tan(x) \cdot \alpha - \frac{1}{2} \tan(x) \ln|1 + \alpha^2| + C \]

### Final Result
Thus, the final result is:
\[ I = \alpha \tan^{-1}(\alpha) \tan(x) - \frac{1}{2} \tan(x) \ln(1 + \alpha^2) + C \]