Integration of rational functions.

Integration of an arbitrary rational function $\frac{P_m(x)}{Q_n(x)}=\frac{b_m{x}^m+...+b_{1}x+b_0}{a_n{x}^n+...+a_{1}x+a_0}$ with real coefficients is performed as follows in the general case.

If $m\ge n$, then it is necessary to preliminarily extract the integer part from this fraction, that is, represent it in the form $$\frac{P_m(x)}{Q_n(x)}=M_{m-n}+\frac{R_r(x)}{Q_n(x)},$$ where $M_{m-n}(x)$ and $R_r(x)$ are polynomials of degrees $m-n\ge 0$ and $r$ respectively, and $r

The extraction of the integer part in the fraction $\frac{P_m(x)}{Q_n(x)}$ is performed by dividing the numerator by the denominator using long division.

To integrate a proper rational function $\frac{P_m(x)}{Q_n(x)},$ $m

Let the denominator $Q_n(x)=a_n{x}^n+...+a_1{x}_1+a_0$ have the decomposition $$Q_n(x)=a_n(x-{\alpha }_1{)}^{s_1}...(x-{\alpha }_l{)}^{s_l}(x^2+p_{1}x+q_1{)}^{t_1}(x^2+p_{k}x+q_k{)}^{t_k}.$$ Then the decomposition of the fraction $\frac{P_m(x)}{Q_n(x)}$ into a sum of simple fractions is as follows:

$$\frac{P_m(x)}{Q_n(x)}=\frac{A_{11}}{x-{\alpha }_1}+...+\frac{A_{s_{1}1}}{(x-{\alpha }_1{)}^{s1}}+...+\frac{A_{1l}}{x-{\alpha }_l}+...+\frac{A_{s_{l}l}}{(x-{\alpha }_l{)}^{s_l}}+$$ $$+\frac{B_{11}x+C_{11}}{x^2+p_{1}x+q_1}+...+\frac{B_{t_{1}1}+C_{t_{1}1}}{(x^2+p_{1}x+q_1{)}^{t_1}}+...$$ $$+\frac{B_{1k}x+C_{1k}}{x^2+p_{k}x+q_k}+...+\frac{B_{t_{k}k}+C_{t_{k}k}}{(x^2+p_{k}x+q_k{)}^{t_k}}.(1)$$

The coefficients $A_{ij},,B_{ij},$ and $C_{ij}$ in this decomposition are determined by equating the coefficients of the same powers of $x$ in the polynomial $P_m(x)$ and the polynomial obtained in the numerator of the right-hand side of equation (1) after bringing it to a common denominator (the method of undetermined coefficients).

Formula (1) shows that integrating an arbitrary rational function reduces to integrating simple fractions of the following four types:

1) $\frac{A}{x-\alpha }.$ $$\int \frac{A}{x-\alpha }dx=A\ln |x-\alpha |+C.$$

2) $\frac{A}{(x-\alpha {)}^k},(k=2,3,...).$

$$\int \frac{A}{(x-\alpha {)}^k}=-\frac{A}{k-1}\frac{1}{(x-\alpha {)}^{k-1}}+C.$$

3) $\frac{Ax+B}{x^2+px+q},p^2-4q<0$

4) $\frac{Ax+B}{(x^2+px+q{)}^k},p^2-4q<0,k=2,3,...$

$\int \frac{2x-1}{x^2+4x+8},dx.$

Solution:

In this example, the discriminant of the quadratic trinomial in the denominator is negative: $p^2-4q=16-32=-16<0,$ which means we have a fraction of the third type. Since $(x^2+4x+8{)}^{\prime }=2x+4,$ we can rewrite the numerator of the fraction as follows: $$2x-1=2x+4-5=(x^2+4x+8{)}^{\prime }-5$$ (this transformation is called extracting the derivative of the quadratic trinomial in the denominator in the numerator). Therefore, $$\int \frac{2x-1}{x^2+4x+8}dx=\int \frac{(x^2+4x+8{)}^{\prime }}{x^2+4x+8}dx-5\int \frac{dx}{x^2+4x+8}.$$

$$\int \frac{(x^2+4x+8{)}^{\prime }}{x^2+4x+8}dx=\left[\begin{array}{l}{x}^2+4x+8=t\\ dt=(x^2+4x+8{)}^{\prime }dx\end{array}\right]=\int \frac{dt}{t}=\ln |t|+C=$$ $$=\ln (x^2+4x+8)+C.$$

We find the integral $\int \frac{dx}{x^2+4x+8}$ by completing the square in the denominator:

$$\int \frac{dx}{x^2+4x+8}=\int \frac{dx}{x^2+4x+4+4}=\int \frac{dx}{(x+2{)}^2+4}=$$ $$=\left[\begin{array}{l}x+2=t\\ dx=dt\end{array}\right]=\int \frac{dt}{t^2+2^2}=\frac{1}{2}arctg\frac{t}{2}+C=\frac{1}{2}arctg\frac{x+2}{2}+C.$$

Thus,

$$\int \frac{2x-1}{x^2+4x+8}dx=\ln (x^2+4x+8)-\frac{5}{2}arctg\frac{x+2}{2}+C.$$

Answer: $\ln (x^2+4x+8)-\frac{5}{2}arctg\frac{x+2}{2}+C.$

2. $\int \frac{3x-1}{(x^2+4x+5{)}^2}.$

Solution.

Here, $p^2-4q=16-20=-4<0,$ which means we have a simple fraction of the fourth type. First, we extract the derivative of the quadratic trinomial in the numerator:

$$(x^2+4x+5{)}^{\prime }=2x+4\Rightarrow 3x-1=\frac{3}{2}(2x+4)-7=$$ $$=1,5(x^2+4x+5{)}^{\prime }-7.$$

$$\int \frac{3x-1}{(x^2+4x+5{)}^2}dx=\int \frac{1,5(x^2+4x+5{)}^{\prime }-7}{(x^2+4x+5{)}^2}dx=$$ $$=\frac{3}{2(x^2+4x+5)}-7\int \frac{dx}{(x^2+4x+5{)}^2}.$$

To compute the remaining integral, we complete the square in the quadratic trinomial: $$\int \frac{dx}{(x^2+4x+5{)}^2}=\int \frac{dx}{((x+2{)}^2+1{)}^2}=\left[\begin{array}{l}x+2=t\\ dx=dt\end{array}\right]=$$ $$=\int \frac{dt}{(t^2+1{)}^2}=\int \frac{1+t^2-t^2}{(1+t^2{)}^2}dt=\int \frac{dt}{1+t^2}-\int \frac{t^{2}dt}{(1+t^2{)}^2}=$$ $$=arctgt-\int \frac{t^{2}dt}{(1+t^2{)}^2}.$$ Then we use integration by parts: $$\int \frac{t^{2}dt}{(t^2+1{)}^2}=\left[\begin{array}{l}u=t\\ dv=\frac{tdt}{(t^2+1{)}^2}\Rightarrow v=-\frac{1}{2}\int d\frac{1}{t^2+1}=-\frac{1}{2(t^2+1)}\end{array}\right]=$$ $$=-\frac{t}{2(1+t^2)}+\frac{1}{2}\int \frac{1}{t^2+1}dt=-\frac{1}{2}(\frac{t}{1+t^2}-arctgt).$$

Thus, $$\int \frac{dx}{(x^2+4x+5{)}^2}=arctg(x+2)+\frac{1}{2}(\frac{x+2}{1+(x+2{)}^2}-arctg(x+2)).$$

And finally, we obtain

$$\int \frac{3x-1}{(x^2+4x+5{)}^2}dx=$$ $$=\frac{3}{2(x^2+4x+5)}+\frac{7}{2}(\frac{x+2}{x^2+4x+5}-3arctg(x+2)).$$

Answer: $\frac{3}{2(x^2+4x+5)}+\frac{7}{2}(\frac{x+2}{x^2+4x+5}-3arctg(x+2)).$

Examples.

Find the integrals:

1. $\int \frac{dx}{x^2+4x-5}.$

Solution.

In this example, $p^2-4q=16+20=36>0,$ which means we will decompose the rational function into simple fractions.

$D=36;$

$$x_1=\frac{-4+6}{2}=1,x_2=\frac{-4-6}{2}=-5.$$

Thus, $$x^2+4x-5=(x-1)(x+5).$$ From this, we have, $$\frac{1}{x^2+4x-5}=\frac{A}{x-1}+\frac{B}{x+5}=\frac{A(x+5)+B(x-1)}{(x-1)(x+5)}=$$ $$=\frac{x(A+B)+(5A-B)}{x^2+4x-5}.$$

By equating the coefficients of the same powers of $x$ in the numerators of the left and right sides of the equation, we write the system:

$$\{\begin{array}{l}A+B=0\\ 5A-B=1\end{array}\Rightarrow \{\begin{array}{l}A=-B\\ -5B-B=1\end{array}\Rightarrow \{\begin{array}{l}A=\frac{1}{6}\\ B=-\frac{1}{6}\end{array}$$

Then the decomposition of the fraction $\frac{1}{x^2+4x-5}$ into a sum of simple fractions is as follows: $$\frac{1}{x^2+4x-5}=\frac{1}{6}\frac{1}{x-1}-\frac{1}{6}\frac{1}{x+5}.$$

Now, let's compute the given integral as the integral of the sum of simple fractions:

$$\int \frac{dx}{x^2+4x-5}=\int (\frac{1}{6}\frac{1}{x-1}-\frac{1}{6}\frac{1}{x+5})dx=$$ $$=\frac{1}{6}\ln |x-1|-\frac{1}{6}\ln |x+5|+C=\frac{1}{6}\ln \left|\frac{x-1}{x+5}\right|+C.$$

Answer: $\frac{1}{6}\ln \left|\frac{x-1}{x+5}\right|+C.$

2.$\int \frac{dx}{x^2-6x}.$

Solution.

$p^2-4q=36>0,$ so we will decompose the rational function into simple fractions.

$$x^2-6x=x(x-6).$$ From this, we have, $$\frac{1}{x^2-6x}=\frac{A}{x}+\frac{B}{x-6}=\frac{A(x-6)+Bx}{x(x-6)}=$$ $$=\frac{x(A+B)-6A}{x^2-6x}.$$

By equating the coefficients of the same powers of $x$ in the numerators of the left and right sides of the equation, we write the system:

$$\{\begin{array}{l}A+B=0\\ -6A=1\end{array}\Rightarrow \{\begin{array}{l}A=-\frac{1}{6}\\ B=\frac{1}{6}\end{array}$$

Thus, the decomposition of the fraction $\frac{1}{x^2-6}$ into a sum of simple fractions is as follows: $$\frac{1}{x^2-6x}=-\frac{1}{6}\frac{1}{x}+\frac{1}{6}\frac{1}{x-6}.$$

Now, let's compute the given integral as the integral of the sum of simple fractions:

$$\int \frac{dx}{x^2-6x}=\int (-\frac{1}{6}\frac{1}{x}+\frac{1}{6}\frac{1}{x-6})dx=$$ $$=-\frac{1}{6}\ln |x|+\frac{1}{6}\ln |x-6|+C=\frac{1}{6}\ln \left|\frac{x-6}{x}\right|+C.$$

Answer: $\frac{1}{6}\ln \left|\frac{x-6}{x}\right|+C.$

3.$\int \frac{xdx}{x^4+6x^2+13}.$

Solution.

Let's make the variable substitution:

$$\int \frac{xdx}{x^4+6x^2+13}=\left[\begin{array}{l}{x}^2=t\\ dt=2xdx\end{array}\right]=\frac{1}{2}\int \frac{dt}{t^2+6t+13}.$$ Here, $p^2-4q=36-52=16<0,$ which means we have a fraction of the third type.

Let's complete the square in the denominator:

$$t^2+6t+13=t^2+6t+9+4=(t+3{)}^2+4.$$

$$\int \frac{dt}{t^2+6t+13}=\int \frac{dt}{(t+3{)}^2+4}=\left[\begin{array}{l}t+3=z\\ dt=dz\end{array}\right]=\int \frac{dz}{z^2+2^2}=$$ $$=\frac{1}{2}arctg\frac{z}{2}+C=\frac{1}{2}arctg\frac{t+3}{2}+C.$$

Thus,

$$\int \frac{xdx}{x^4+6x^2+13}dx=\frac{1}{4}arctg\frac{x^2+3}{2}+C.$$

Answer: $\frac{1}{4}arctg\frac{x^2+3}{2}+C.$

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

Solution.

The integrand has the form $\frac{P_n(x)}{Q_n(x)},$ where $P_n(x)$ and $Q_n(x)$ are polynomials of degree $n=3,$ which means the given fraction is improper. Let's extract the integer part from this fraction. After division in a column of $x^3+2$ by $x^3-4,$ we obtain $$\frac{x^3+2}{x^3-4x}=1+\frac{4x+2}{x^3-4x}.$$ Next, we decompose the obtained fraction into a sum of simple fractions:

$$\frac{4x+2}{x^3-4x}=\frac{4x+2}{x(x^2-4)}=\frac{4x+2}{x(x-2)(x+2)}=\frac{A}{x}+\frac{B}{x-2}+\frac{C}{x+2}=$$ $$=\frac{A(x^2-4)+B(x^2+2x)+C(x^2-2x)}{x^3-4x}=$$ $$=\frac{x^2(A+B+C)+x(2B-2C)-4A}{x^3-4x}.$$

By equating coefficients of the same degrees of $x$ in the numerators of the left and right sides of the equation, we write the system:

$$\{\begin{array}{l}A+B+C=0\\ 2B-2C=4\\ -4A=2\end{array}\Rightarrow \{\begin{array}{l}B+C=\frac{1}{2}\\ B-C=2\\ A=-\frac{1}{2}\end{array}\Rightarrow \{\begin{array}{l}B=\frac{1}{2}-C\\ \frac{1}{2}-C-C=2\\ A=-\frac{1}{2}\end{array}\Rightarrow$$ $$\Rightarrow \{\begin{array}{l}B=\frac{5}{4}\\ C=-\frac{3}{4}\\ A=-\frac{1}{2}\end{array}$$

Thus, the decomposition of the fraction $\frac{4x+2}{x^3-4x}$ into a sum of simple fractions is given by $$\frac{4x+2}{x^3-4x}=-\frac{1}{2}\frac{1}{x}+\frac{5}{4}\frac{1}{x-2}-\frac{3}{4}\frac{1}{x+2}.$$

Next, let's compute the given integral as the integral of the sum of simple fractions:

$$\int \frac{x^3+2}{x^3-4x}dx=\int (1+\frac{4x+2}{x^3-4x})dx=$$ $$=\int (-\frac{1}{2}\frac{1}{x}+\frac{5}{4}\frac{1}{x-2}-\frac{3}{4}\frac{1}{x+2})dx=$$ $$=x-\frac{1}{2}\ln |x|+\frac{5}{4}\ln |x-2|-\frac{3}{4}\ln |x+2|+C.$$

Answer: $x-\frac{1}{2}\ln |x|+\frac{5}{4}\ln |x-2|-\frac{3}{4}\ln |x+2|+C.$

5. $\int \frac{(x-1)dx}{(x^2+1{)}^3}.$

Solution.

Here, $p^2-4q=-4<0$, which means we have a simple fraction of the fourth type. First, we extract the derivative of the quadratic trinomial from the numerator:

$$(x^2+1{)}^{\prime }=2x\Rightarrow x-1=\frac{1}{2}(2x)-1=\frac{1}{2}(x^2+1{)}^{\prime }-1.$$

$$\int \frac{x-1}{(x^2+1{)}^3}dx=\int \frac{0,5(x^2+1{)}^{\prime }-1}{(x^2+1{)}^3}dx=$$ $$=0,5\int (x^2+1{)}^{-3}d(x^2+1)-\int \frac{dx}{(x^2+1{)}^3}=$$ $$=-\frac{1}{4(x^2+1{)}^2}-\int \frac{dx}{(x^2+1{)}^3}.$$

Let's compute the remaining integral:

$$\int \frac{dx}{(x^2+1{)}^3}=\int \frac{1+x^2-x^2}{(1+x^2{)}^3}dx=\int \frac{dx}{(1+x^2{)}^2}-\int \frac{x^{2}dx}{(1+x^2{)}^3}=$$ $$=\int \frac{1+x^2-x^2}{(1+x^2{)}^2}dx-\int \frac{x^{2}dx}{(1+x^2{)}^3}=$$ $$=\int \frac{dx}{1+x^2}-\int \frac{x^{2}dx}{(1+x^2{)}^2}-\int \frac{x^{2}dx}{(1+x^2{)}^3}=$$ $$=arctgx-\int \frac{x^{2}dx}{(1+x^2{)}^2}-\int \frac{x^{2}dx}{(1+x^2{)}^3}.$$

Next, we will use the method of integration by parts:

$$\int \frac{x^{2}dx}{(x^2+1{)}^2}=\left[\begin{array}{l}u=x\\ dv=\frac{xdx}{(x^2+1{)}^2}\Rightarrow v=-\frac{1}{2}\int d\frac{1}{x^2+1}=\frac{-1}{2(x^2+1)}\end{array}\right]=$$ $$=-\frac{x}{2(1+x^2)}+\frac{1}{2}\int \frac{1}{x^2+1}dt=\frac{1}{2}(arctgx-\frac{x}{1+x^2}).$$

$$\int \frac{x^{2}dx}{(x^2+1{)}^3}=$$ $$\left[\begin{array}{l}u=x\\ dv=\frac{xdx}{(x^2+1{)}^3}\Rightarrow v=-\frac{1}{4}\int d\frac{1}{(x^2+1{)}^2}=-\frac{1}{4(x^2+1{)}^2}\end{array}\right]=$$ $$-\frac{x}{4(1+x^2{)}^2}+\frac{1}{4}\int \frac{1}{(x^2+1{)}^2}dx.$$

$$\int \frac{1}{(x^2+1{)}^2}dx=arctgx-\int \frac{x^2}{(1+x^2{)}^2}dx=$$ $$=arctgx-\frac{1}{2}(arctgx-\frac{x}{1+x^2})=\frac{1}{2}(\frac{x}{1+x^2}+arctgx).$$

Thus, $$\int \frac{x^{2}dx}{(x^2+1{)}^3}=-\frac{x}{4(1+x^2{)}^2}+\frac{1}{8}(\frac{x}{1+x^2}+arctgx).$$

And finally, we get

$$\int \frac{x-1}{(x^2+1{)}^3}dx=$$ $$-\frac{1}{4(x^2+1{)}^2}-(arctgx-\frac{1}{2}(arctgx-\frac{x}{1+x^2})-$$ $$-(-\frac{x}{4(x^2+1{)}^2}+\frac{1}{8}(\frac{x}{1+x^2}+arctgx)))=+C$$ $$=-\frac{x+1}{4(x^2+1{)}^2}-\frac{3}{8}arctgx-\frac{3}{8}\frac{x}{1+x^2}+C.$$

Answer: $-\frac{x+1}{4(x^2+1{)}^2}-\frac{3}{8}arctgx-\frac{3}{8}\frac{x}{1+x^2}+C.$

Homework.

Find the integrals.

1. $\int \frac{dx}{2x^2-4x+5}.$

2. $\int \frac{xdx}{x^2-5x+4}.$

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

4. $\int \frac{2x^2-1}{x^3-5x^2+6x}dx.$

59. $\int \frac{x^4+3x^3+3x^2-5}{x^3+3x^2+3x+1}dx.$

6. $\int \frac{3x^2+2x-1}{(x-1{)}^2(x+2)}dx.$

7. $\int \frac{5x-13}{(x^2-5x+6{)}^2}dx.$

Tags: Integration of rational functions, calculus, integral, mathematical analysis