Evaluate $\int\frac{\mathrm{d}x}{{(x^4+2x+10)}^4}$ I am having trouble with this integral  $\displaystyle \int \frac{\mathrm{d}x}{{(x^4+2x+10)}^4}$
I know this can be solved using the "ostrogradsky" method. But that is too lengthy.
i tried using integration by parts by multiplying and dividing by $4x^3+2$ but nothing popped up .
Could anyone put me on the right track.
 A: In the same spirit as @lab bhattacharjee, because of the fourth power in denominatour, assume that $$\int\frac{dx}{{(x^4+2x+10)}^4}=\frac{P_n(x)}{{(x^4+2x+10)}^3}$$ Differentiate both sides and remove the denominator to get
$$0=-1+\left(x^4+2 x+10\right) P_n'(x)-6 \left(2 x^3+1\right) P_n(x)\tag 1$$ For me, this differential equation does not admit a polynomial solution because of the $1$.
Trying with
$$P_6(x)=a+b x+c x^2+d x^3+e x^4+f x^5+g x^6$$ and expanding $(1)$, we then have
$$(-6 a+10 b-1)+x (20 c-4 b)+x^2 (30 d-2 c)+x^3 (40 e-12 a)+x^4 (-11 b+2 e+50
   f)+x^5 (-10 c+4 f+60 g)+x^6 (6 g-9 d)-8 e x^7-7 f x^8-6 g
   x^9$$ Starting from the end, the last  terms give $e=f=g=0$; since $g=0$ then $d=0$; then  $c=0$; then $b=0$ and finally $a=0$ and what remains if the $-1$ !
So, and I am ready to bet that there is a typo in the problem, the solution could be to consider
$$\frac{1}{{(x^4+2x+10)}^4}=\frac 1 {(x-a)^4(x-b)^4(x-c)^4(x-d)^4}$$ where $a,b,c,d$ are the complex roots of $x^4+2x+10=0$ and use partial fraction decomposition (this would be a pure nightmare) to end with a bunch of integrals looking like
$$I=\int \frac {dx} {(x-k)^n} \qquad \text{with} \qquad n=1,2,3,4\qquad \text{and} \qquad k=\text{complex number}$$
