# Integrals of simple fractions

A type of integral that we may find are those integrals of a polynomial fraction.

$$\displaystyle \int \frac{x+4}{x^2-5x+3} \ dx$$

In a more general way, the integrals of the kind$$\displaystyle \int R(x) \ dx= \int \frac{P(x)}{F(x)}$$, where $$P(x)$$ and $$F(x)$$ are polynomials.

In the case in which degree $$P(x) \geqslant$$ degree $$F(x)$$, it is necessary to do the polynomials' division to obtain:

$$\displaystyle\frac{P(x)}{F(x)}=Q(x)+\frac{f(x)}{F(x)}$$, where degree $$f(x))<$$ degree $$F(x)$$, and then we do the decomposition into simple fractions of $$\displaystyle\frac{f(x)}{F(x)}$$.

## Decomposition in simple fractions

In order to decompose a polynomial fraction into simple fractions, we first have to express the denominator as a product of polynomials of 1st and 2nd degree.

Then, we equalize the function to a sum of terms: given $$\displaystyle\frac{f(x)}{F(x)}$$, we decompose, as we have seen in the polynomials topic, $$F(x)$$ as a product of polynomials of degree 1 and 2: $$F(x)=a_{m}x^m+a_{m-1}x^{m-1}+\ldots+a_{2}x^2+a_{1}x+a_{0}=$$$$$=a_{m} \cdot (x-a)^{\alpha} \cdot (x-b)^{\beta}\cdot \ldots (x^2+ox+q)^\rho\cdot(x^2+rx+s)^\lambda$$$ , where $$a$$, $$b$$, etc are the roots of the polynomial, of multiplicity $$\alpha, \beta, \ldots$$, and $$p$$, $$q$$, $$r$$, $$s$$ are the coefficients of the factors of order $$2$$.

In this fashion, we take the following equality:$$\displaystyle\frac{f(x)}{F(x)}=\frac{A_1}{x-a}+\frac{A_2}{(x-a^2)}+ \ldots+\frac{A_{\alpha}}{(x-a)^\alpha}+\frac{B_1}{x-b}+\frac{B_2}{(x-b)^2}+$$$$$+\ldots+\frac{B_\beta}{(x-b)^\beta}+\frac{M_1x+N_1}{x^2+px+q}+\frac{M_2x+N_2}{(x^2+px+q)^2}+$$$ $$+\ldots+\frac{M_px+N_p}{(x^2+px+q)^p}+\ldots$$$where $$A$$'s,$$B$$'s, $$M$$'s and $$N$$'s are unknown. Next, we do the sum of all these polynomial functions, using the common denominator, and we equalize this to the initial polynomial fraction, equating the coefficients of every degree of the numerator. As soon as the coefficients are obtained, we can express the original integral as a sum of integrals that we know how to solve using the logarithm an the arctangent. ### Procedure to follow 1. Make sure that the degree of the numerator is greater than the one of the denominator. If that is not the case, simplify the fraction by doing the polynomials' division. 2. Decompose the polynomial in the denominator, by Ruffini or any other method. 3. Write the polynomial fraction as a sum of fractions as has been described previously, obtaining several unknown constants 4. Extract common factors of the denominators, and obtain a system of equations obtained by equating the terms of the same degree. 5. Solve the system of equations. 6. Re-write the original integral as the sum of integrals of fractions of degree 1 or 2, and solve the integral, bearing in mind that:$$\displaystyle \int \frac{1}{x+a} \ dx=\ln|x+a|+C \\ \displaystyle \int \frac{1}{x^2+a^2} \ dx=\frac{1}{a} \arctan \Big(\frac{x}{a}\Big)+C$$$ $$\displaystyle \int \frac{x}{x^2+a^2} \ dx=\frac{1}{2}\ln|x^2+a^2|+C$$$$$\displaystyle \int \frac{x-2}{(x-1)^2\cdot(x^2+1)} \ dx$$ We have, in this case, $$\displaystyle \frac{x-2}{(x-1)^2\cdot(x^2+1)}=\frac{A_1}{x-1}+\frac{A_2}{(x-1)^2}+\frac{M_1·x+N_1}{x^2+1}=$$$ $$=\dfrac{A_1(x^2+1)(x-1)+A_2(x^2+1)+M_1x(x-1)^2+N_1(x-1)^2}{(x-1)^2(x^2+1)}$$$We obtain the following system of equations: $$\begin{array} {ll} 0=A_1+M_1 \\ 0=A_2-A_1-2\cdot M_1+N_1 \\ 1=A_1+M_1-2 N_1\\-2=A_2-A_1+N_1\end{array}$$$ Solving it we obtain $$\begin{array}{ll} A_1=1 \\ A_2=-\frac{1}{2} \\ M_1=-1 \\ N_1=-\frac{1}{2}\end{array}$$$And, therefore, $$\displaystyle \int \frac{x-2}{(x-1)^2\cdot(x^2+1)} \ dx = \int \frac{1}{x-1} \ dx - \frac{1}{2} \int \frac{1}{(x-1)^2} \ dx + \int \frac{-x-\frac{1}{2}}{x^2+1} \ dx=$$$ $$= \ln|x-1|+\frac{1}{6}(x-1)^{-3}-\displaystyle \int \frac{x}{x^2+1} \ dx - \frac{1}{2} \int \frac{1}{x^2+1} \ dx =$$$$$= \ln|x-1|+\frac{1}{6}(x-1)^{-3}-\frac{1}{2}\ln|x^2+1|-\frac{1}{2}\arctan x+C$$$

$$\displaystyle \int \frac{x^4-4x^2+x+1}{x^3+x^2-4x-4} \ dx$$

We will show the process step by step:

• The degree of the numerator is greater than the one of the denominator, we then have to do the polynomial division, obtaining the result:

$$\displaystyle\frac{x^4-4x^2+x+1}{x^3+x^2-4x-4}=x-1+\frac{x^2+x-3}{x^3+x^2-4x-4}$$

so that

$$\displaystyle \int \frac{x^4-4x^2+x+1}{x^3+x^2-4x-4} \ dx = \int x-1 ++\frac{x^2+x-3}{x^3+x^2-4x-4} \ dx=$$

$$=\frac{x^2}{2}-x+\int \frac{x^2+x-3}{x^3+x^2-4x-4} \ dx$$

and we will compute:

$$\displaystyle\int \frac{x^2+x-3}{x^3+x^2-4x-4} \ dx$$

• Decomposing, we have: $$x^3+x^2-4x-4=(x+2)(x-2)(x+1)$$

$$\displaystyle \frac{x^2+x-3}{x^3+x^2-4x-4}=\frac{A_1}{x+2}+\frac{A_2}{x-2}+\frac{A_3}{x+1}=$$

$$= \displaystyle \frac{A_1(x-2)(x+1)}{x^3+x^2-4x-4}+\frac{A_2(x+2)(x+1)}{x^3+x^2-4x-4}+\frac{A_3(x-2)(x+2)}{x^3+x^2-4x-4}$$

$$\displaystyle\frac{x^2+x-3}{x^3+x^2-4x-4}=\frac{A_1(x^2-x-2)}{x^3+x^2-4x-4}+\frac{A_2(x^2+3x+2}{x^3+x^2-4x-4}+\frac{A_3(x^2-4)}{x^3+x^2-4x-4}$$

and, therefore, we have:

$$\begin{array}{l} 1=A_1+A_2+A_3 \\ 1=-A_1+3A_2 \\ -3=-2A_1+2A_2-4A_3 \end{array}$$

• From the previous system of equations, we obtain: $$\begin{array} {l} A_1=\frac{-1}{4} \\ A_2=\frac{1}{4} \\ A_3=1 \end{array}$$

$$\displaystyle \int \frac{x^4-4x^2+x+1}{x^3+x^2-4x-4} \ dx = \frac{x^2}{2}-x+\int \frac{\frac{-1}{4}}{x+2} \ dx +\int \frac{\frac{1}{4}}{x-2} \ dx +\int \frac{1}{x+1} \ dx=$$

$$=\displaystyle\frac{x^2}{2}-x-\frac{1}{4}\ln|x+2|+\frac{1}{4}\ln|x-2|+\ln|x+1| +C$$