If `int(xtan^(-1))/(sqrt(1+x^(2)))dx=`
`sqrt(1+x^(2)) f(x)+Klog(x+sqrt(x^(2)+1))+C` then

Options

A. `f(x)=tan^(-1)x,K=-1`
B. `f(x)=tan^(-1)x,K=1`
C. `f(x)=2tan^(-1)x,K=-1`
D. `f(x)=2tan^(-1)x,K=1`

Answer

To solve the integral \( I = \int \frac{x \tan^{-1}(x)}{\sqrt{1+x^2}} \, dx \), we will use a substitution method and integration by parts.

### Step 1: Substitution
Let \( x = \tan(\theta) \). Then, we have:
\[ dx = \sec^2(\theta) \, d\theta \]
Also, we know that:
\[ \sqrt{1+x^2} = \sqrt{1+\tan^2(\theta)} = \sec(\theta) \]
Thus, the integral becomes:
\[ I = \int \frac{\tan(\theta) \theta}{\sec(\theta)} \sec^2(\theta) \, d\theta \]
This simplifies to:
\[ I = \int \tan(\theta) \theta \sec(\theta) \, d\theta \]

### Step 2: Integration by Parts
Now, we will use integration by parts. Let:
- \( u = \theta \) (which gives \( du = d\theta \))
- \( dv = \tan(\theta) \sec(\theta) \, d\theta \)

To find \( v \), we know that:
\[ \int \tan(\theta) \sec(\theta) \, d\theta = \sec(\theta) + C \]
Thus, \( v = \sec(\theta) \).

Now applying integration by parts:
\[ I = u v - \int v \, du \]
Substituting the values:
\[ I = \theta \sec(\theta) - \int \sec(\theta) \, d\theta \]

### Step 3: Integral of Secant
The integral of \( \sec(\theta) \) is:
\[ \int \sec(\theta) \, d\theta = \log |\sec(\theta) + \tan(\theta)| + C \]
Thus, we have:
\[ I = \theta \sec(\theta) - \log |\sec(\theta) + \tan(\theta)| + C \]

### Step 4: Back Substitution
Now we substitute back \( \theta = \tan^{-1}(x) \) and \( \sec(\theta) = \sqrt{1+x^2} \):
\[ I = \tan^{-1}(x) \sqrt{1+x^2} - \log |\sqrt{1+x^2} + x| + C \]

### Step 5: Comparison
We need to compare this with the form:
\[ \sqrt{1+x^2} f(x) + K \log(x + \sqrt{x^2 + 1}) + C \]
From our result:
\[ I = \tan^{-1}(x) \sqrt{1+x^2} - \log |\sqrt{1+x^2} + x| + C \]
We can see that:
- \( f(x) = \tan^{-1}(x) \)
- \( K = -1 \)

### Final Result
Thus, we conclude that:
\[ f(x) = \tan^{-1}(x) \quad \text{and} \quad K = -1 \]