[Cambridge/N78/I/20] 求 $\frac{\text{d}}{\text{d}x}(\tan^{-1}(x^2))$，并由此或用其他方法，求 $\int x \tan^{-1}(x^2) \, dx$。

1. Find the derivative of $\tan^{-1}(x^2)$ using the chain rule.

$\frac{\text{d}}{\text{d}x} (\tan^{-1}(x^2)) = \frac{1}{1+(x^2)^2} \cdot \frac{\text{d}}{\text{d}x}(x^2)$
$= \frac{2x}{1+x^4}$

2. Set up integration by parts for the given integral.

To find $\int x \tan^{-1}(x^2) \, dx$, use integration by parts $\int u \frac{\text{d}v}{\text{d}x} \, dx = uv - \int v \frac{\text{d}u}{\text{d}x} \, dx$:
Let $u = \tan^{-1}(x^2) \implies \frac{\text{d}u}{\text{d}x} = \frac{2x}{1+x^4}$
Let $\frac{\text{d}v}{\text{d}x} = x \implies v = \frac{1}{2}x^2$

3. Apply the integration by parts formula.

$\int x \tan^{-1}(x^2) \, dx = \left( \tan^{-1}(x^2) \right) \left( \frac{1}{2}x^2 \right) - \int \left( \frac{1}{2}x^2 \right) \left( \frac{2x}{1+x^4} \right) \, dx$
$= \frac{1}{2} x^2 \tan^{-1}(x^2) - \int \frac{x^3}{1+x^4} \, dx$

4. Evaluate the remaining integral using a standard integration form.

Notice that the numerator $x^3$ is a scalar multiple of the derivative of the denominator $1+x^4$.
$\int \frac{x^3}{1+x^4} \, dx = \frac{1}{4} \int \frac{4x^3}{1+x^4} \, dx = \frac{1}{4} \ln(1+x^4)$
Thus,
$\int x \tan^{-1}(x^2) \, dx = \frac{1}{2} x^2 \tan^{-1}(x^2) - \frac{1}{4} \ln(1+x^4) + C$