How do you solve this differential equation using variation of parameters? $\color{green}{question}$:
How do you solve this differential equation using variation of parameters?

$$y"-\frac{2x}{x^2+1}y'+\frac{2}{x^2+1}y=6(x^2+1)$$

$\color{green}{I~tried}$ . . .
$using~the~\color{blue}{Laplace~transform}~method$ . . .
$$L[\int_{0}^{\infty }\frac{sinxt}{1+t^{2}}dt]$$
$$=\int_{0}^{\infty }e^{-sx}(\int_{0}^{\infty }\frac{sinxt}{1+t^{2}}dt)$$
$$=\int_{0}^{\infty }\frac{1}{1+t^{2}}(\int_{0}^{\infty }e^{-px}sinxtdx)dt\\\\\\=\int_{0}^{\infty }\frac{1}{1+t^{2}}\frac{t}{s^{2}+t^{2}}dt$$
$$=\int_{0 }^{\infty }\frac{1}{s^{2}-1}(\frac{t}{1+t^{2}}-\frac{t}{s^{2}+t^{2}})$$
$$=\frac{Lns}{s^{2}-1}$$
Is my solution correct?
Should I use the inverse Laplace?
How can I get a complete and correct answer?
Thanks for any hint.

I will map out the steps for you and have you fill in the details.
We are asked to solve this using Variation of Parameters (VoP), given:
$$\tag 1 y''-\dfrac{2x}{x^2+1}y'+\dfrac{2}{x^2+1}y=6(x^2+1)$$
Step 1
Find the homogeneous solution to $(1)$, so we have:
$$\tag 2 y''-\dfrac{2x}{x^2+1}y'+\dfrac{2}{x^2+1}y=0$$
This yields:
$$y_h = c_1(x^2-1) + c_2 x$$
Step 2
We are now going to make use of VoP, so we set: $y_1 = x^2-1$ and $y_2 = x$ from $y_h$ and $f = 6(x^2+1)$ from $(1)$.
We calculate the Wronskian of $y_1$ and $y_2$, yielding $W(x^2-1, x) = -x^2-1$.
Using VoP, we have:
$$u_1 = \int \dfrac{-y_2 f}{W(x^2-1, x)} dx = \int \dfrac{-x 6(x^2+1)}{-x^2-1} dx = 3x^2$$
$$u_2 = \int \dfrac{y_1 f}{W(x^2-1, x)} dx = \int \dfrac{(x^2-1)6(x^2+1)}{-x^2-1} dx = 6x-2x^3$$
Now, $y_p$ is given by:
$$y_p = y_1 u_1 + y_2 u_2 = (x^2-1)(3x^2) + (-2x^3+6x)(x) = x^4 + 3x^2$$
Step 3
Our final solution is given by:
$$y(x) = y_h(x) + y_p(x) = c_1(x^2-1) + c_2 x + x^4 + 3x^2$$
