How can one integrate $\displaystyle\int\frac{1}{(x+1)^4(x^2+1)}\ dx$?
Attempt:
I tried partial fraction decomposition (PFD) and got lost. The method of u-substitution didn't work for me either.
What else can I do? Can one calculate the integral without PFD?
Best Answer
Here is a secure and faster method when the fraction has a pole of comparatively high order:
$$\frac 1{u^4(2-2u+u^2)}=\frac1{2u^4}+\frac 1{2u^3} u+\frac 1{4u^2}-\frac1{4(2-2u+u^2)},$$ or with $x$ : $$\frac 1{(x+1)^4(x^2+1)}=\frac1{2(x+1)^4}+\frac 1{2(x+1)^3} +\frac 1{4(x+1)^2}-\frac1{4(x^2+1)}.$$