`int(x^2tan^(-1)x)/(1+x^2)dx =`

Options

A. `x tan^(-1)x-1/2log(1+x^2)-1/3(tan^(-1)x)^2`
B. `x tan^(-1)x+1/2(1+x^2)+1/2(tan^(-1)x)^2`
C. `x tan^(-1)x-1/2log(1+x^2)+1/2(tan^(-1)x)^2`
D. None of these

Answer

To solve the integral

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

we can use integration by parts and some algebraic manipulation. Let's break it down step by step.

### Step 1: Rewrite the integral
We can rewrite the integrand by separating the terms. We notice that we can express \(x^2\) as \( (1 + x^2) - 1 \):

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

This simplifies to:

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

### Step 2: Integrate the first part
Now we need to integrate \( \int \tan^{-1}(x) \, dx \). We will use integration by parts here. Let:

- \( u = \tan^{-1}(x) \) → \( du = \frac{1}{1+x^2} \, dx \)
- \( dv = dx \) → \( v = x \)

Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \):

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

### Step 3: Simplify the second integral
Now we need to compute \( \int \frac{x}{1+x^2} \, dx \). We can use a simple substitution:

Let \( w = 1 + x^2 \) → \( dw = 2x \, dx \) → \( dx = \frac{dw}{2x} \).

This gives us:

\[ \int \frac{x}{1+x^2} \, dx = \frac{1}{2} \int \frac{1}{w} \, dw = \frac{1}{2} \ln |w| + C = \frac{1}{2} \ln(1+x^2) + C. \]

### Step 4: Combine the results
Now substituting back, we have:

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

### Step 5: Integrate the second part
Now we need to compute \( -\int \frac{\tan^{-1}(x)}{1+x^2} \, dx \). This integral can be solved using the formula for the integral of \( \frac{f'(x)}{f(x)} \):

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

### Final Result
Combining everything, we get:

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