If `int (xtan^(-1)x)/sqrt(1+x^2) dx = sqrt(1+x^2)f(x)+Aln|x+sqrt(x^2+1)|+c` then

Options

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

Answer

To solve the integral

\[ \int \frac{x \tan^{-1}(x)}{\sqrt{1+x^2}} \, dx \]

we will use integration by parts and a substitution method. Let's denote

\[ I = \int \frac{x \tan^{-1}(x)}{\sqrt{1+x^2}} \, dx. \]

### Step 1: Integration by Parts
We will use integration by parts, where we let:
- \( u = \tan^{-1}(x) \)
- \( dv = \frac{x}{\sqrt{1+x^2}} \, dx \)

Now, we need to find \( du \) and \( v \):
- The derivative of \( u \) is:
\[ du = \frac{1}{1+x^2} \, dx \]
- To find \( v \), we integrate \( dv \):
\[ v = \int \frac{x}{\sqrt{1+x^2}} \, dx. \]
Using the substitution \( w = 1+x^2 \), we have \( dw = 2x \, dx \), or \( dx = \frac{dw}{2x} \). This gives us:
\[ v = \int \frac{x}{\sqrt{w}} \cdot \frac{dw}{2x} = \frac{1}{2} \int w^{-1/2} \, dw = \sqrt{1+x^2}/2. \]

### Step 2: Applying Integration by Parts
Now, applying the integration by parts formula \( \int u \, dv = uv - \int v \, du \):
\[ I = \left(\tan^{-1}(x) \cdot \frac{\sqrt{1+x^2}}{2}\right) - \int \frac{\sqrt{1+x^2}}{2} \cdot \frac{1}{1+x^2} \, dx. \]

### Step 3: Simplifying the Integral
The remaining integral is:
\[ \int \frac{\sqrt{1+x^2}}{2(1+x^2)} \, dx = \frac{1}{2} \int \frac{1}{\sqrt{1+x^2}} \, dx. \]
This integral can be solved using the substitution \( x = \tan(\theta) \), leading to:
\[ \int \frac{1}{\sqrt{1+x^2}} \, dx = \ln |x + \sqrt{1+x^2}| + C. \]

### Step 4: Final Expression
Putting it all together, we have:
\[ I = \frac{\tan^{-1}(x) \sqrt{1+x^2}}{2} - \frac{1}{2} \ln |x + \sqrt{1+x^2}| + C. \]

### Conclusion
Thus, we can express the integral as:
\[ \int \frac{x \tan^{-1}(x)}{\sqrt{1+x^2}} \, dx = \sqrt{1+x^2} \cdot f(x) + A \ln |x + \sqrt{1+x^2}| + C, \]
where \( f(x) = \frac{1}{2} \tan^{-1}(x) \) and \( A = -\frac{1}{2} \).