Integral of $\frac{(x^2+1)}{(x^2-2)(x+1)^3}$?

Question

I was trying to do integral of $$\frac{x^2+1}{(x^2-2)(x+1)^3}$$ starting off by simplifying the equation into something like: $$\frac{A}{(x^2-2)} + \frac{Bx + C}{(x+1)^3}$$ and then determine A, B and C.

I notice that $(x+1)^3$ can still be done by breaking the $(x+1)^3$ down itself, but when it is something like by the power of 4 or 5 or something beyond, it becomes too complicated to be done.

I would like to ask if there is any easier way for me to break the equation down and do the integral. Thanks

Answer 1

Using partial fraction decomposition, just write $$\frac{x^2+1}{(x^2-2)(x+1)^3}=\frac A{x+1}+\frac B{(x+1)^2}+\frac C{(x+1)^3}+\frac {Dx+E}{x^2-2}$$ Reduce to common denominator to get $$x^2+1=A(x+1)^2(x^2-2)+B(x+1)(x^2-2)+C(x^2-2)+(Dx+E)(x+1)^3$$ Expand and identify the coefficients.

A bit tedious but not difficult.

Answer 2

You should write $\dfrac{x^2+1}{(x^2-2)(x+1)^3}$as$$\frac A{x+\sqrt2}+\frac B{x-\sqrt2}+\frac C{x+1}+\frac D{(x+1)^2}+\frac E{(x+1)^3},$$which is equivalent to solving the system$$\left\{\begin{array}{l}-\sqrt2A+\sqrt2B-2C-2D-2E=1\\-3\sqrt2A+A+3\sqrt2B+B-4C-2D=0\\-3\sqrt2A+3A+3\sqrt2B+3B-C+D+E=1\\-\sqrt{2}A+3A+\sqrt2B+3B+2C+D=0\\A+B+C=0.\end{array}\right.$$It turns out the the solution is$$(A,B,C,D,E)=\left(\frac34\left(10+7\sqrt2\right),-\frac34\left(-10+7\sqrt2\right),-15,6,-2\right).$$Can you take it from here?