Elementary Calculus

$\int\frac{x(x^2-1)}{x^6+1}dt$ ---> assume : $x^2=a$ ---> $2xdx=da$

---> $\frac{1}{2}\int\frac{a-1}{a^3+1} da=\frac{1}{2}\int\frac{a-1}{(a+1)(a^2-a+1)} da$

---> $\frac{a-1}{(a+1)(a^2-a+1)}=\frac{P}{a+1}-\frac{Qa+R}{a^2-a+1}$

$a-1=(P-Q)a^2+(-P-Q-R)a+(P-R)$ ---> $P=Q=-\frac{2}{3}, R=\frac{1}{3}$

---> $\frac{1}{2}\int\frac{a-1}{(a+1)(a^2-a+1)} da = \frac{1}{6}\int(\frac{-2}{a+1}+\frac{2a-1}{a^2-a+1})da$

We must evaluate the integral of $\int\frac{2a-1}{a^2-a+1}da$ first :

---> $\int\frac{2a-1}{a^2-a+1}da = 4\int\frac{2a-1}{(2a-1)^2+3} da$

---> consider : $\sqrt[2]{3}tanθ=2a-1$ ---> $a=\frac{\sqrt[2]{3}tanθ+1}{2}$ ---> $da=\frac{\sqrt[2]{3}}{2}sec^2θ dθ$

$2\int\frac{\sqrt[2]{3}.tanθ}{3.sec^2θ}\sqrt[2]{3}sec^2θ dθ = 2\int\tanθ dθ = -2.ln|cosθ| = -2ln|\frac{\sqrt[2]{3}}{\sqrt[2]{4a^2-4a+4}}|+ C$

Then, we have :

$\frac{1}{6}\int(\frac{-2}{a+1}+\frac{2a-1}{a^2-a+1})da = \frac{1}{6}(-2.ln|a+1|-2ln|\frac{\sqrt[2]{3}}{\sqrt[2]{4a^2-4a+4}}|+ C) = \frac{1}{3}.ln|\frac{\sqrt[2]{4x^4-4x^2+4}}{\sqrt[2]{3}(x^2+1)}|+K$ , $K=\frac{1}{6}C$

---> I.e the desired value is $4-4+4+1=5$

$\begin{aligned} I & = \int \frac {x^3-x}{x^6+1} dx \\ & = \int \frac {x^3-x}{(x^2+1)(x^4-x^2+1)} dx & \small \color{#3D99F6} \text{By partial fraction decomposition} \\ & = \int \left(\frac {-2x}{3(x^2+1)} + \frac {2x^3-x}{3(x^4-x^2+1)} \right) dx \\ & = \int \left(\frac {-2x}{3(x^2+1)} + {\color{#3D99F6} \frac {2x^3-x}{3(x^2+\sqrt 3x+1)(x^2-\sqrt 3x+1)}} \right) dx & \small \color{#3D99F6} \text{By partial fraction decomposition again} \\ & = \int \left(\frac {-2x}{3(x^2+1)} + {\color{#3D99F6}\frac {2x+\sqrt 3}{6(x^2+\sqrt 3x+1)} +\frac {2x-\sqrt 3}{6(x^2-\sqrt 3x+1)}}\right) dx \\ & = - \frac {\ln(x^2+1)}3 + \frac {\ln(x^2+\sqrt 3x+1)}6 + \frac {\ln(x^2-\sqrt 3x+1)}6 + \color{#3D99F6}C & \small \color{#3D99F6} \text{where }C \text{ is the arbitrary constant of integration.} \\ & = \frac 13 \ln \left(\frac {\sqrt{x^4-x^2+1}}{x^2+1}\right) + C \end{aligned}$

Therefore, $a+b+c+d = 1+1+1+1 = \boxed{4}$