To solve the integral $\int \frac{3x^2 \tan^{-1}(x)}{1+x^2} \, dx$ without trigonometric substitution or the beta function, we can use integration by parts.

Let:

• $u = \tan^{-1}(x)$ ⇒ $du = \frac{1}{1+x^2} \, dx$
• $dv = 3x^2 \, dx$ ⇒ $v = x^3$

Now, apply the integration by parts formula $\int u \, dv = uv - \int v \, du$:

$\int \frac{3x^2 \tan^{-1}(x)}{1+x^2} \, dx = x^3 \tan^{-1}(x) - \int x^3 \cdot \frac{1}{1+x^2} \, dx$

Next, simplify $\int \frac{x^3}{1+x^2} \, dx$:

$\frac{x^3}{1+x^2} = x - \frac{x}{1+x^2}$

So, the integral becomes:

$\int x \, dx - \int \frac{x}{1+x^2} \, dx$

Calculating these integrals:

1. $\int x \, dx = \frac{x^2}{2}$
2. For $\int \frac{x}{1+x^2} \, dx$, let $w = 1+x^2$, then $dw = 2x \, dx$:

$\int \frac{x}{1+x^2} \, dx = \frac{1}{2} \ln|1+x^2|$

Now combine everything:

$\int \frac{3x^2 \tan^{-1}(x)}{1+x^2} \, dx = x^3 \tan^{-1}(x) - \left( \frac{x^2}{2} - \frac{1}{2} \ln|1+x^2| \right) + C$

Thus, the final result is:

$\int \frac{3x^2 \tan^{-1}(x)}{1+x^2} \, dx = x^3 \tan^{-1}(x) - \frac{x^2}{2} + \frac{1}{2} \ln(1+x^2) + C$

Would you like more details or do you have any questions? Here are some related questions you might find interesting:

1. How can integration by parts be applied in other scenarios?
2. What are the geometric interpretations of the functions involved?
3. Can this approach be generalized to other forms of integrals?
4. What are the limitations of integration by parts?
5. How does this result compare to numerical integration methods?

Tip: Always keep track of your constants during integration to avoid losing terms!