Integrate it - 2

Let $I=\displaystyle \int^{1}_{{1/ \sqrt3}}{\dfrac{x^2+3}{x^6\left(x^2+1 \right)}\mathrm{d}x }$

$\Rightarrow I=\displaystyle \int^{1}_{{1/\sqrt3}}{\dfrac{\left(x^2+1 \right) +2}{x^6\left(x^2+1 \right)}\mathrm{d}x= \int^{1}_{{1/ \sqrt3}}{\dfrac{1}{x^6}+\dfrac{2}{x^6\left(x^2+1 \right)}}}\mathrm{d}x$

Using partial fraction decomposition in the second part and adding the first term to it,we finally get

$\displaystyle\int^{1}_{{1/\sqrt3}}{-\dfrac{2}{x^2+1}+\dfrac{2}{x^2}-\dfrac{2}{x^4}+\dfrac{3}{x^6}\,\mathrm{d}x}$ $\Rightarrow I= \left[-2\arctan\left(x\right)-\dfrac{2}{x}+\dfrac{2}{3x^3}-\dfrac{3}{5x^5}\right]_{1/\sqrt3}^{1}$ $\Rightarrow I=\left[-2\arctan\left(x\right)-\dfrac{2}{x}+\dfrac{2}{3x^3}-\dfrac{3}{5x^5} \right]_{1/\sqrt3}^{1} =\dfrac{5{\pi}+3^{9/2}}{15}-\dfrac{15{\pi}+58}{30}=\boxed{6.89614}$

For variety, I am showing another way to solve the integral. Since the nominator $P (x)$ and denominator $Q (x)$ of the integrand are polynomials with real coefficients and $P (x) / Q (x)$ is a proper fraction, and $Q (x)$ has multiple roots, we can use the Ostrogradsky method which gives the following.

$\int \frac {P(x)}{Q(x)} dx = \frac {P_1(x)}{Q_1(x)} + \int \frac {P_2(x)}{Q_2(x)} dx$

where $Q_1(x)$ is the greatest common divisor of $Q (x)$ and its derivative $Q' (x)$ , while $Q_2(x) = Q (x) / Q_1(x)$ .

As $Q(x) = x^6(x^2+1)$ , $Q'(x) = 6x^5(x^2+1) + 2x^7$ , the greatest common divisor $Q_1(x) = x^5$ and $Q_2 = x(x^2+1)$ . Let $P_1(x) = Ax^4+Bx^3+Cx^2+Dx+E$ and $P_2(x) = Fx^2+Gx + H$ . Then, we have:

$\begin{aligned} \int_{\frac 1{\sqrt 3}}^1 \frac {x^2+3}{x^6(x^2+1)} dx & = \frac {Ax^4+Bx^3+Cx^2+Dx+E}{x^5} + \int \frac {Fx^2+Gx + H}{x(x^2+1)} dx \\ \frac {x^2+3}{x^6(x^2+1)} & = \frac {(4Ax^3+3Bx^2+2Cx+D)x-5(Ax^4+Bx^3+Cx^2+Dx+E)}{x^6} + \frac {Fx^2+Gx + H}{x(x^2+1)} \\ x^2+3 & = \left((4Ax^3+3Bx^2+2Cx+D)x-5(Ax^4+Bx^3+Cx^2+Dx+E)\right)(x^2+1) + (Fx^2+Gx + H)x^5 \\ & = \left(-Ax^4-2Bx^3-3Cx^2-4Dx-5E\right)(x^2+1) + Fx^7+Gx^6 + Hx^5 \\ & = Fx^7 + (G-A)x^6 + (H-2B)x^5- (A+3C)x^4-(2B+4D)x^3-(3C+5E)x^2-4Dx-5E \end{aligned}$

Equating the coefficients of $x^n$ on both sides, we get $A=-2$ , $B=0$ , $C= \frac 23$ , $D=0$ , $E=-\frac 35$ , $F=0$ , $G=-2$ and $H=0$ . Then, we have:

$\begin{aligned} \int_{\frac 1{\sqrt 3}}^1 \frac {x^2+3}{x^6(x^2+1)} dx & = \frac {-2x^4+\frac 23 x^2 - \frac 35}{x^5} \bigg|_{\frac 1{\sqrt 3}}^1+ \int_{\frac 1{\sqrt 3}}^1 \frac {-2x}{x(x^2+1)} dx \\ & = \frac {-30x^4+10 x^2 - 9}{15x^5} \bigg|_{\frac 1{\sqrt 3}}^1- 2\int_{\frac 1{\sqrt 3}}^1 \frac 1{x^2+1} dx \\ & =-\frac {29}{15}+\frac {27\sqrt 3}5- 2\tan^{-1} x \bigg|_{\frac 1{\sqrt 3}}^1 \\ & =-\frac {29}{15}+\frac {27\sqrt 3}5 - \frac \pi 2 + \frac \pi 3 \\ & \approx \boxed{6.896} \end{aligned}$