Question: User: \int \frac{x^2(1+\sin(x)\cos(x))^2}{(x^2+1)^3(\sin(x)+\cos(x))^4 \ln(x^2+1)} \, dx
1. **Stating the problem:** We want to evaluate the integral
$$\int \frac{x^2(1+\sin(x)\cos(x))^2}{(x^2+1)^3(\sin(x)+\cos(x))^4 \ln(x^2+1)} \, dx$$
2. **Analyzing the integral:** The integrand is quite complex, involving powers of $x^2+1$, trigonometric sums and products, and a logarithm $\ln(x^2+1)$.
3. **Look for substitutions:** Notice the denominator has $(x^2+1)^3$ and $\ln(x^2+1)$, suggesting a substitution involving $u = \ln(x^2+1)$ or $u = x^2+1$.
4. **Try substitution $u = x^2 + 1$:**
Then $du = 2x \, dx$ or $x \, dx = \frac{du}{2}$.
But the numerator has $x^2$, so express $x^2 = u - 1$.
Rewrite the integral in terms of $u$:
$$\int \frac{(u-1)(1+\sin(x)\cos(x))^2}{u^3 (\sin(x)+\cos(x))^4 \ln(u)} \, dx$$
But $dx$ is not directly expressed in $du$; we have $x \, dx = \frac{du}{2}$, so $dx = \frac{du}{2x}$.
This substitution is complicated by the trigonometric terms in $x$.
5. **Try substitution $v = \sin(x) + \cos(x)$:**
Note that $v = \sin(x) + \cos(x)$, then
$$dv = \cos(x) - \sin(x) \, dx$$
But the integral has $(\sin(x) + \cos(x))^4$ and $(1 + \sin(x)\cos(x))^2$.
Rewrite $1 + \sin(x)\cos(x)$:
Recall $\sin(x)\cos(x) = \frac{1}{2} \sin(2x)$, so
$$1 + \sin(x)\cos(x) = 1 + \frac{1}{2} \sin(2x)$$
This is complicated to substitute directly.
6. **Try rewriting numerator and denominator:**
Rewrite numerator:
$$(1 + \sin(x)\cos(x))^2 = \left(1 + \frac{1}{2} \sin(2x)\right)^2 = 1 + \sin(2x) + \frac{1}{4} \sin^2(2x)$$
Rewrite denominator:
$$(\sin(x) + \cos(x))^4 = (\sqrt{2} \sin(x + \frac{\pi}{4}))^4 = 4 \sin^4(x + \frac{\pi}{4})$$
7. **Due to complexity, no elementary antiderivative is apparent.**
8. **Conclusion:** The integral does not simplify easily with elementary functions and standard substitutions.
It likely requires advanced special functions or numerical methods.
**Final answer:** The integral cannot be expressed in elementary closed form with standard methods.