求 $\int \frac{\cos x}{\cos 3x + \cos x} \text{ d}x$。 求 $\int x \tan^{-1} x^2 \text{ d}x$。由此，求 $\int_{-1}^{1} |x \tan^{-1} x^2| \text{ d}x$ 的精确值。

1. Simplify the denominator using the factor formula to cancel the common term with the numerator.

$\cos 3x + \cos x = 2\cos\left(\frac{3x+x}{2}\right)\cos\left(\frac{3x-x}{2}\right)$
$= 2\cos 2x \cos x$
$\frac{\cos x}{\cos 3x + \cos x} = \frac{\cos x}{2\cos 2x \cos x} = \frac{1}{2\cos 2x} = \frac{1}{2}\sec 2x$

2. Integrate the simplified expression using the standard result for the secant function from the formula list.

$\int \frac{\cos x}{\cos 3x + \cos x} \text{ d}x = \int \frac{1}{2}\sec 2x \text{ d}x$
$= \frac{1}{2} \left( \frac{1}{2} \ln |\sec 2x + \tan 2x| \right) + C$
$= \frac{1}{4} \ln |\sec 2x + \tan 2x| + C$

1. Use integration by parts to integrate the function.

Let $u = \tan^{-1} x^2$ and $\frac{\text{d}v}{\text{d}x} = x$.
Then $\frac{\text{d}u}{\text{d}x} = \frac{2x}{1+(x^2)^2} = \frac{2x}{1+x^4}$ and $v = \frac{1}{2}x^2$.
Using the integration by parts formula $\int u \frac{\text{d}v}{\text{d}x} \text{ d}x = uv - \int v \frac{\text{d}u}{\text{d}x} \text{ d}x$,
$\int x \tan^{-1} x^2 \text{ d}x = \frac{1}{2}x^2 \tan^{-1} x^2 - \int \frac{1}{2}x^2 \left( \frac{2x}{1+x^4} \right) \text{ d}x$
$= \frac{1}{2}x^2 \tan^{-1} x^2 - \int \frac{x^3}{1+x^4} \text{ d}x$

2. Integrate the remaining algebraic fraction.

Notice that the numerator is a scalar multiple of the derivative of the denominator. Let $f(x) = 1+x^4$, then $f'(x) = 4x^3$.
$\int \frac{x^3}{1+x^4} \text{ d}x = \frac{1}{4} \int \frac{4x^3}{1+x^4} \text{ d}x = \frac{1}{4} \ln(1+x^4)$
So, the indefinite integral is:
$\int x \tan^{-1} x^2 \text{ d}x = \frac{1}{2}x^2 \tan^{-1} x^2 - \frac{1}{4} \ln(1+x^4) + C$

3. Simplify the definite integral by exploiting the symmetry of the integrand.

Let $g(x) = |x \tan^{-1} x^2|$.
Testing for symmetry: $g(-x) = |-x \tan^{-1} (-x)^2| = |-x| \tan^{-1} x^2 = |x| \tan^{-1} x^2 = g(x)$.
Since $g(x)$ is an even function,
$\int_{-1}^{1} |x \tan^{-1} x^2| \text{ d}x = 2 \int_{0}^{1} |x \tan^{-1} x^2| \text{ d}x$
For $x \in [0, 1]$, $x \ge 0$ and $\tan^{-1} x^2 \ge 0$, so $|x \tan^{-1} x^2| = x \tan^{-1} x^2$.
Thus, the integral simplifies to:
$2 \int_{0}^{1} x \tan^{-1} x^2 \text{ d}x$

4. Evaluate the simplified definite integral to find the exact value.

Using the result from the indefinite integral:
$2 \int_{0}^{1} x \tan^{-1} x^2 \text{ d}x = 2 \left[ \frac{1}{2}x^2 \tan^{-1} x^2 - \frac{1}{4} \ln(1+x^4) \right]_{0}^{1}$
Substitute the limits:
$= 2 \left( \left( \frac{1}{2}(1^2)\tan^{-1}(1^2) - \frac{1}{4} \ln(1+1^4) \right) - \left( 0 - \frac{1}{4} \ln(1) \right) \right)$
$= 2 \left( \frac{1}{2} \left(\frac{\pi}{4}\right) - \frac{1}{4} \ln 2 - 0 \right)$
$= 2 \left( \frac{\pi}{8} - \frac{1}{4} \ln 2 \right)$
$= \frac{\pi}{4} - \frac{1}{2} \ln 2$