Evaluate the following integrals:
`int(x^(2)tan^(-1)x)/((1+x^(2)))dx`

Explanation

To evaluate the integral

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

we can use integration by parts and some algebraic manipulation. Here’s a step-by-step solution:

### Step 1: Rewrite the Integral
We can rewrite the integral by separating \(x^2\) into \(x^2 + 1 - 1\):

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

This can be split into two separate integrals:

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

### Step 2: Evaluate the First Integral
The first integral is straightforward:

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

Using integration by parts, let \(u = \tan^{-1}(x)\) and \(dv = dx\). Then \(du = \frac{1}{1+x^2} \, dx\) and \(v = x\).

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

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

### Step 3: Evaluate the Second Integral
Now we need to evaluate

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

This can be solved using a simple substitution. Let \(w = 1 + x^2\), then \(dw = 2x \, dx\) or \(\frac{1}{2} dw = x \, dx\).

Thus,

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

### Step 4: Combine the Results
Now we can combine the results from Steps 2 and 3:

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

### Step 5: Substitute Back into the Original Integral
Now substituting back into our original expression:

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

### Step 6: Final Result
Thus, the final result is:

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