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Evaluate the integral of x4 / (x4 +5x2 + 4)

by RoRi

Compute the following integral.

    \[ \int \frac{x^4 \, dx}{x^4 + 5x^2 + 4}. \]

First, we simplify the integrand and then use partial fractions,

    \[ \int \frac{x^4 \, dx}{x^4 + 5x^2 + 4} = \int \frac{x^4 + 5x^2 + 4 - 5x^2 - 4}{x^4 + 5x^2 + 4} \, dx = \int \, dx - \int \frac{5x^2 + 4}{x^4 + 5x^2 + 4} \, dx. \]

To evaluate the second integral we use partial fractions. We have

    \[ x^4 + 5x^2 + 4 = (x^2+4)(x^2+1). \]

Therefore, we write

    \[ \frac{5x^2+4}{x^4 + 5x^2 + 4} = \frac{Ax+B}{x^2+4} + \frac{Cx + D}{x^2+1}. \]

Then we have the equation

    \begin{align*} &&(Ax+B)(x^2+1) + (Cx+D)(x^2+4) &= 5x^2 + 4 \\ \implies && (A+C)x^3 + (B+D)x^2 + (A+4C)x + (B+4D) &= 5x^2 + 4. \end{align*}

Equating like powers of x we then have four equations and four unknowns:

    \begin{align*} A+C &= 0 \\ B+D &= 5 \\ A+4C &= 0 \\ B+4D &= 4. \end{align*}

Solving this system we find A = C = 0, B = \frac{16}{3} and D = -\frac{1}{3}. Therefore, we have

    \begin{align*} \int \frac{x^4 \, dx}{x^4 + 5x^2 + 4} &= \int dx - \int \frac{5x^2+4}{(x^2+4)(x^2+1)} \, dx \\ &= x - \frac{16}{3} \int \frac{dx}{x^2+4} + \frac{1}{3} \int \frac{dx}{1+x^2} \\ &= x + \frac{1}{3} \arctan x - \frac{8}{3} \arctan \left( \frac{x}{2} \right) + C. \end{align*}

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