Integration Using Partial Fractions

When integrating a rational function with a denominator of degree 2 or higher, we can often use partial fractions to simplify the expression into a sum of integrals that can be easily evaluated.

Partial fractions are a technique used to decompose a rational function into simpler fractions. The method relies on the fact that any proper rational function can be expressed as a sum of simpler rational functions whose denominators are linear or irreducible quadratic factors.

The Partial Fractions Decomposition

Suppose we have a rational function:

P(x)Q(x)=a1xz1+a2xz2++an(xzn)k+b1x+c1x2+p1x+q1++bmx+cmx2+pmx+qm\frac{P(x)}{Q(x)}=\frac{a_1}{x-z_1}+\frac{a_2}{x-z_2}+\cdots+\frac{a_n}{(x-z_n)^k}+\frac{b_1x+c_1}{x^2+p_1x+q_1}+\cdots+\frac{b_mx+c_m}{x^2+p_mx+q_m}

where P(x)P(x) and Q(x)Q(x) are polynomials with Q(x)Q(x) of degree n2n\geq2, and z1,,zn,p1,,pmz_1, \ldots, z_n, p_1, \ldots, p_m are distinct complex numbers.

The partial fractions decomposition of P(x)Q(x)\frac{P(x)}{Q(x)} is the expression on the right-hand side of the equation above. The coefficients a1,,an,b1,,bm,c1,,cma_1, \ldots, a_n, b_1, \ldots, b_m, c_1, \ldots, c_m are constants that can be determined by equating the numerator of the original rational function to the numerator of the partial fraction decomposition and simplifying.

Example

Let's consider the rational function:

3x2+5x+2x3+3x2+2x\frac{3x^2+5x+2}{x^3+3x^2+2x}

The denominator is a cubic polynomial, so it is difficult to integrate directly. However, we can use partial fractions to simplify the expression.

First, we factor the denominator:

x3+3x2+2x=x(x2+3x+2)=x(x+1)(x+2)x^3+3x^2+2x=x(x^2+3x+2)=x(x+1)(x+2)

Now we can write:

3x2+5x+2x3+3x2+2x=Ax+Bx+1+Cx+2\frac{3x^2+5x+2}{x^3+3x^2+2x}=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{x+2}

Multiplying both sides by the denominator gives:

3x2+5x+2=A(x+1)(x+2)+Bx(x+2)+Cx(x+1)3x^2+5x+2=A(x+1)(x+2)+Bx(x+2)+Cx(x+1)

Setting x=0x=0 and solving for AA gives A=1A=1. Setting x=1x=-1 and solving for BB gives B=2B=2. Setting x=2x=-2 and solving for CC gives C=1C=-1.

Therefore, we have:

3x2+5x+2x3+3x2+2x=1x+2x+11x+2\frac{3x^2+5x+2}{x^3+3x^2+2x}=\frac{1}{x}+\frac{2}{x+1}-\frac{1}{x+2}

Now we can integrate each of the simpler fractions separately:

1xdx=lnx+C12x+1dx=2lnx+1+C21x+2dx=lnx+2+C3\begin{aligned} \int\frac{1}{x}dx&=\ln|x|+C_1 \\ \int\frac{2}{x+1}dx&=2\ln|x+1|+C_2 \\ \int\frac{-1}{x+2}dx&=-\ln|x+2|+C_3 \\ \end{aligned}

where C1,C2,C_1, C_2, and C3C_3 are constants of integration.

Therefore, the original integral is:

3x2+5x+2x3+3x2+2xdx=lnx+2lnx+1lnx+2+C\int\frac{3x^2+5x+2}{x^3+3x^2+2x}dx=\ln|x|+2\ln|x+1|-\ln|x+2|+C

where C=C1+C2+C3C=C_1+C_2+C_3 is the constant of integration.

Conclusion

Partial fractions are a powerful tool for simplifying the integration of rational functions with difficult denominators. By decomposing a rational function into a sum of simpler fractions, we can easily evaluate each term and then sum them up to obtain the integral of the original function. It is important to note that partial fractions will only work if the denominator is a product of linear or irreducible quadratic factors.

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