Let us consider the integral $\int\frac{5x-3}{x^2-2x-3}dx$. $\frac{d}{dx}(x^2-2x-3)=2x-2$ so substitution is not an option. Noting $x^2-2x-3=(x+1)(x-3)$, let us assume instead that $$\frac{5x-3}{x^2-2x-3}=\frac{A}{x+1}+\frac{B}{x-3}$$ Then \begin{align*}5x-3&=A(x-3)+B(x+1)\\&=(A+B)x-3A+B\end{align*} Hence we obtain a system of linear equations $$\left\{\begin{aligned}A+B&=5\\-3A+B&=-3\end{aligned}\right.$$ Solving this system simultaneously we find $A=2$ and $B=3$.

Alternation: Let’s begin with $5x-3=A(x-3)+B(x+1)$. For $x=3$, we get $12=4B$ so $B=3$. For $x=-1$, we get $-8=-4A$ so $A=2$. This method certainly has a computational advantage when it works over getting a system of linear equations and solving it.

Now \begin{align*}\int\frac{5x-3}{x^2-2x-3}dx&=2\int\frac{dx}{x+1}+3\int\frac{dx}{x-3}\\&=2\ln|x+1|+3\ln|x-3|+C\end{align*}

General method of writing a rational function $\frac{f(x)}{g(x)}$ as a sum of partial fractions

1. Let $x-r$ be a linear factor of $g(x)$. Suppose that $(x-r)^m$ is the highest power of $x-r$ that divides $g(x)$ i.e. $x=r$ is a zero of $g(x)$ with multiplicity $m$. Then to this factor, assign the sum of the $m$ partial fractions $$\frac{A_1}{x-r}+\frac{A_2}{(x-r)^2}+\cdots+\frac{A_m}{(x-r)^m}$$
2. Let $x^2+px+q$ be a quadratic factor of $g(x)$ that cannot be factored further into linear factors with real coefficients. Suppose that $(x^2+px+q)^n$ is the highest power of this factor that divides $g(x)$. Then to this factor assign the sum of the $n$ partial fractions $$\frac{B_1x+C_1}{x^2+px+q}+\frac{B_2x+C_2}{(x^2+px+q)^2}+\cdots+\frac{B_nx+C_n}{(x^2+px+q)^n}$$

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

Solution. Let $$\frac{x^2+4x+1}{(x-1)(x+1)(x+3)}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{x+3}$$ Then $$A(x+1)(x+3)+B(x-1)(x+3)+C(x-1)(x+1)$$ For $x=1$, $8A=6$ so $A=\frac{3}{4}$. For $x=-1$, $-4B=-2$ so $B=\frac{1}{2}$. For $x=-3$, $8C=-2$ so $C=-\frac{1}{4}$. Hence, \begin{align*}\int\frac{x^2+4x+1}{(x-1)(x+1)(x+3)}dx&=\frac{3}{4}\int\frac{dx}{x-1}+\frac{1}{2}\int\frac{dx}{x+1}-\frac{1}{4}\int\frac{dx}{x+3}\\&=\frac{3}{4}\ln|x-1|+\frac{1}{2}\ln|x+1|-\frac{1}{4}\ln|x+3|+C\end{align*}

Example. Evaluate $\int\frac{6x+7}{(x+2)^2}dx$.

Solution. Let $\frac{6x+7}{(x+2)^2}=\frac{A}{x+2}+\frac{B}{(x+2)^2}$. Then $A=6$ and $B=-5$. Hence, \begin{align*}\int\frac{6x+7}{(x+2)^2}dx\\&=6\int\frac{dx}{x+2}-5\int\frac{dx}{(x+2)^2}\\&=6\ln|x+2|+\frac{5}{x+2}+C\end{align*}

Example. [Integrating an Improper Fraction] Evaluate $\int\frac{2x^3-4x^2-x-3}{x^2-2x-3}dx$.

Solution. By long division, divide $2x^3-4x^2-x-3$ by $x^2-2x-3$ to obtain quotient $2x$ and remainder $5x-3$. Thus $2x^3-4x^2-x-3=(x^2-2x-3)\cdot 2x+5x-3$ and $$\frac{2x^3-4x^2-x-3}{x^2-2x-3}=2x+\frac{5x-3}{x^2-2x-3}$$ $\frac{5x-3}{x^2-2x-3}$ is a proper fraction so we can apply the above method to write it as $$\frac{5x-3}{x^2-2x-3}=\frac{3}{x-3}+\frac{2}{x+1}$$ Hence, \begin{align*}\int\frac{2x^3-4x^2-x-3}{x^2-2x-3}dx&=\int 2xdx+3\int\frac{dx}{x-3}+2\int\frac{dx}{x+1}\\&=x^2+3\ln|x-3|+2\ln|x+1|+C\end{align*}

Remark. Instead of writing $\frac{5x-3}{x^2-2x-3}$ as $\frac{A}{x-3}+\frac{B}{x+1}$, let $$5x-3=A(x+1)+B(x-3)$$ $\frac{5x-3}{x+1}=A+\frac{B(x-3)}{x+1}$ and if $x=3$, $A=3$. $\frac{5x-3}{x-3}=\frac{A(x+1)}{x-3}+B$ and if $x=-1$, $B=2$. This is called Heaviside’s method and is easier to determine coefficients than the standard method we discussed above. However it can be useful only when the denominator has all linear factors.

Example. Evaluate $\int\frac{-2x+4}{(x^2+1)(x-1)^2}dx$.

Solution. Let $$\frac{-2x+4}{(x^2+1)(x-1)^2}=\frac{Ax+B}{x^2+1}+\frac{C}{x-1}+\frac{D}{(x-1)^2}$$ Then we obtain \begin{align*}-2x+4&=(Ax+B)(x-1)^2+C(x-1)(x^2+1)+D(x^2+1)\\&=(A+C)x^3+(-2A+B-C+D)x^2+(A-2B+C)x+(B-C+D)\end{align*} and hence the equations $A+C=0$, $-2A+B-C+D=0$, $A-2B+C=-2$, and $B-C+D=4$. Solve these equations simultaneously to obtain $A=2$, $B=1$, $C=-2$, and $D=1$. Therefore, \begin{align*}\int\frac{-2x+4}{(x^2+1)(x-1)^2}dx&=\int\frac{2x+1}{x^2+1}dx-2\int\frac{dx}{x-1}+\int\frac{dx}{(x-1)^2}\\&=\int\frac{2x}{x^2+1}dx+\int\frac{1}{x^2+1}dx-2\int{dx}{x-1}+\int\frac{dx}{(x-1)^2}\\&=\ln(x^2+1)+\tan^{-1}x-2\ln|x-1|-\frac{1}{x-1}+C\end{align*}

Example. Evaluate $\int\frac{dx}{x(x^2+1)^2}$.

Solution. Let $$\frac{1}{x(x^2+1)^2}=\frac{A}{x}+\frac{Bx+C}{x^2+1}+\frac{Dx+E}{(x^2+1)^2}$$ Then we have \begin{align*}1&=A(x^2+1)^2+(Bx+C)x(x^2+1)+(Dx+E)x\\&=(A+B)x^4+Cx^3+(2A+B+D)x^2+(C+E)x+A\end{align*} and comparing the coefficients we obtain the equations $A+B=0$, $C=0$, $2A+B+D=0$, $C+E=0$, and $A=1$. Solve these equations simultaneously to obtain $A=1$, $B=-1$, $C=0$, $D=-1$, and $E=0$. Therefore, \begin{align*}\int\frac{dx}{x(x^2+1)^2}&=\int\frac{dx}{x}-\int\frac{x}{x^2+1}dx-\int\frac{x}{(x^2+1)^2}dx\\&=\ln|x|-\frac{1}{2}\ln(x^2+1)+\frac{1}{2(x^2+1)}+C\end{align*}