Question

What is the integration of x^4/[{x^2+a^2}{x^2+b^2}]?

Answer 1

Answer on Question #51705 – Math - Integral Calculus

What is the integration of $x^4 / \left\{ x^2 + a^2 \right\} \left\{ x^2 + b^2 \right\}$?

Solution.

$$I = \int \frac {x ^ {4} d x}{(x ^ {2} + a ^ {2}) (x ^ {2} + b ^ {2})}$$

Step 1. Let's consider the integrand, i.e. the function that is to be integrated, $f(x)$:

$$\begin{array}{l} f (x) = \frac {x ^ {4}}{(x ^ {2} + a ^ {2}) (x ^ {2} + b ^ {2})} = \frac {x ^ {4}}{x ^ {4} + x ^ {2} (a ^ {2} + b ^ {2}) + a ^ {2} b ^ {2}} = \\ = \frac {x ^ {4} + \left(x ^ {2} \left(a ^ {2} + b ^ {2}\right) + a ^ {2} b ^ {2}\right) - \left(x ^ {2} \left(a ^ {2} + b ^ {2}\right) + a ^ {2} b ^ {2}\right)}{x ^ {4} + x ^ {2} \left(a ^ {2} + b ^ {2}\right) + a ^ {2} b ^ {2}} = \\ = 1 - \frac {x ^ {2} \left(a ^ {2} + b ^ {2}\right) + a ^ {2} b ^ {2}}{x ^ {4} + x ^ {2} \left(a ^ {2} + b ^ {2}\right) + a ^ {2} b ^ {2}} = 1 - \frac {x ^ {2} \left(a ^ {2} + b ^ {2}\right) + a ^ {2} b ^ {2}}{\left(x ^ {2} + a ^ {2}\right) \left(x ^ {2} + b ^ {2}\right)} \\ \end{array}$$

Step 2. Let's use the method of Partial Fraction Decomposition.

$$\begin{array}{l} \frac {x ^ {2} \left(a ^ {2} + b ^ {2}\right) + a ^ {2} b ^ {2}}{\left(x ^ {2} + a ^ {2}\right) \left(x ^ {2} + b ^ {2}\right)} = \frac {A x + B}{\left(x ^ {2} + a ^ {2}\right)} + \frac {C x + D}{\left(x ^ {2} + b ^ {2}\right)} = \\ = \frac {A x \left(x ^ {2} + b ^ {2}\right) + B \left(x ^ {2} + b ^ {2}\right) + C x \left(x ^ {2} + a ^ {2}\right) + D \left(x ^ {2} + a ^ {2}\right)}{\left(x ^ {2} + a ^ {2}\right) \left(x ^ {2} + b ^ {2}\right)} = \\ = \frac {A x ^ {3} + B x ^ {2} + A x b ^ {2} + B b ^ {2} + C x ^ {3} + D x ^ {2} + C x a ^ {2} + D a ^ {2}}{\left(x ^ {2} + a ^ {2}\right) \left(x ^ {2} + b ^ {2}\right)} = \\ = \frac {x ^ {3} (A + C) + x ^ {2} (B + D) + x (A b ^ {2} + C a ^ {2}) + B b ^ {2} + D a ^ {2}}{(x ^ {2} + a ^ {2}) (x ^ {2} + b ^ {2})} \\ \end{array}$$

Step 3. Equate the left-hand and the right-hand sides, multiply through by the denominator so we no longer deal with fractions.

$$x ^ {2} \left(a ^ {2} + b ^ {2}\right) + a ^ {2} b ^ {2} = x ^ {3} (A + C) + x ^ {2} (B + D) + x \left(A b ^ {2} + C a ^ {2}\right) + B b ^ {2} + D a ^ {2}$$

Step 4. Find unknown coefficients $A, B, C$ and $D$, equating like powers of $x$ in the last line. Now we have the system of four equations and solving them gives us unknown coefficients $A, B, C$ and $D$.

$$\begin{array}{l} x^{3}: \quad A + C = 0 \quad A = -C \quad A = 0 \\ x^{2}: \quad B + D = a^{2} + b^{2} \quad \Rightarrow \quad Bb^{2} + Db^{2} = a^{2}b^{2} + b^{4} \quad C = 0 \\ x^{1}: \quad Ab^{2} + Ca^{2} = 0 \quad \Rightarrow \quad A = -\frac{Ca^{2}}{b^{2}} \quad \Rightarrow \quad Bb^{2} + Db^{2} = a^{2}b^{2} + b^{4} \quad \Rightarrow \\ x^{0}: \quad Bb^{2} + Da^{2} = a^{2}b^{2} \quad \quad \quad Bb^{2} + Da^{2} = a^{2}b^{2} \quad \quad \quad Bb^{2} + Da^{2} = a^{2}b^{2} \\ A = 0 \\ C = 0 \\ \Rightarrow D = \frac{b^{4}}{b^{2} - a^{2}} \\ B = -\frac{a^{4}}{b^{2} - a^{2}} \\ \end{array}$$

Step 5. Rewrite the integrand substituting the unknown coefficients $A, B, C$ and $D$:

$$\begin{array}{l} f(x) = 1 - \frac{x^{2}(a^{2} + b^{2}) + a^{2}b^{2}}{(x^{2} + a^{2})(x^{2} + b^{2})} = 1 - \frac{Ax + B}{(x^{2} + a^{2})} - \frac{Cx + D}{(x^{2} + b^{2})} = \\ = 1 - \frac{B}{(x^{2} + a^{2})} - \frac{D}{(x^{2} + b^{2})} = 1 + \frac{a^{4}}{(b^{2} - a^{2})} \frac{1}{(x^{2} + a^{2})} - \frac{b^{4}}{(b^{2} - a^{2})} \frac{1}{(x^{2} + b^{2})}. \\ \end{array}$$

Step 6. Compute the initial indefinite Integral:

$$\begin{array}{l} I = \int \frac{x^{4}dx}{(x^{2} + a^{2})(x^{2} + b^{2})} = \int dx \left(1 + \frac{a^{4}}{(b^{2} - a^{2})} \frac{1}{(x^{2} + a^{2})} - \frac{b^{4}}{(b^{2} - a^{2})} \frac{1}{(x^{2} + b^{2})}\right) = \\ = x + \frac{a^{4}}{(b^{2} - a^{2})} \frac{1}{a} \arctan \frac{x}{a} - \frac{b^{4}}{(b^{2} - a^{2})} \frac{1}{b} \arctan \frac{x}{b} + c = \\ = x + \frac{a^{3}}{(b^{2} - a^{2})} \arctan \frac{x}{a} - \frac{b^{3}}{(b^{2} - a^{2})} \arctan \frac{x}{b} + c, \\ \end{array}$$

where $c$ is an integration constant.

Answer: $I = \int \frac{x^4 dx}{(x^2 + a^2)(x^2 + b^2)} = x + \frac{a^3}{(b^2 - a^2)} \arctan \frac{x}{a} - \frac{b^3}{(b^2 - a^2)} \arctan \frac{x}{b} + c.$

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