Integration of algebraic fractions

To integrate a fraction of polynomials one proceeds in two stages: first the degree of the numerator is lowered, then the second-degree denominator is handled by distinguishing three cases according to the sign of the discriminant $\Delta$.

Stage 1. If $\deg\text{num} \ge \deg\text{den}$, one carries out polynomial division to reduce to a numerator of lower degree.

Stage 2. For $\displaystyle\int\frac{dx+e}{ax^2+bx+c}\,dx$ three cases are distinguished:

• Case A ($\Delta>0$): one factorises $a(x-x_1)(x-x_2)$ and decomposes into partial fractions $\dfrac{C_1}{x-x_1}+\dfrac{C_2}{x-x_2}$. Practical trick: “I cover $(x-x_1)$ with my hand and substitute $x=x_1$” to find $C_1$.
• Case B ($\Delta=0$): one uses the substitution $t=x-x_1$.
• Case C ($\Delta<0$): one completes the square in the denominator, reducing to an $\arctan$.

Example — Case A: $\int_3^5\dfrac{2x+3}{5(x-1)(x+2)}\,dx$

With the “hand” trick: $C_1 = \frac{2\cdot 1+3}{5(1+2)} = \frac{1}{3}, \qquad C_2 = \frac{2(-2)+3}{5(-2-1)} = \frac{1}{15}.$ Hence $\int_3^5\frac{2x+3}{5(x-1)(x+2)}\,dx = \left[\frac{1}{3}\ln|x-1|+\frac{1}{15}\ln|x+2|\right]_3^5.$

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Topics: Integrale
Concepts: Completamento del quadrato · Frazioni parziali
Methods: Integrale frazioni parziali
Skills: Integrare · Ragionare per casi · Scomporre