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Evaluate the integral of (x4 + 1) / (x (x2 + 1)2)

by RoRi

Compute the following integral.

    \[ \int \frac{x^4 + 1}{x(x^2+1)^2} \, dx.\]

The denominator is already factored, so we write

    \[ \frac{x^4+1}{x(x^2+1)^2} = \frac{A}{x} + \frac{Bx+C}{x^2+1} + \frac{Dx+E}{(x^2+1)^2}. \]

This gives us the equation

    \[ A(x^2+1)^2 + (Bx+C)(x)(x^2+1) + (Dx+E)x = x^4 + 1. \]

First, we can evaluate at x = 0 to obtain

    \[ A = 1. \]

Then, we multiply out and equate like powers of x,

    \begin{align*} &&(x^2+1)^2 + (Bx+C)(x)(x^2+1) + (Dx+E)x &= x^4 + 1 \\ \implies && x^4 + 2x^2 + 1 + Bx^4 + Cx^3 + Bx^2 + Cx + Dx^2 + Ex &= x^4 + 1 \\ \implies && (1+B)x^4 + Cx^3 + (2 + B + D)x^2 + (C + E)x &= x^4. \end{align*}

Therefore, equating like powers of x, we have the following equations

    \begin{align*} 1+B &= 1 \\ C &= 0 \\ 2+B+D &= 0 \\ C+E &= 0 \end{align*}

Solving this system we obtain

    \[ B = 0, \quad C = 0, \quad D = -2, \quad E = 0. \]

Therefore, evaluating the integral we have

    \begin{align*} \int \frac{x^4+1}{x(x^2+1)^2} \, dx &= \int \left( \frac{A}{x} + \frac{Bx+C}{x^2+1} + \frac{Dx+E}{(x^2+1)^2} \right) \, dx \\[9pt] &= \int \frac{1}{x} \, dx - \int \frac{2x}{(x^2+1)^2} \, dx \\[9pt] &= \log |x| + \frac{1}{x^2+1} + C. \end{align*}

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