Calculus An intuitive approach

Let us find the partial fraction decomposition of the rational function $F(x)={\displaystyle \frac{x-1}{2x^2-x-3}}\text{.}$ The denominator $Q(x)=2x^2-x-3$ of $F(x)$ is reducible since its discriminant $(-1{)}^2-4(2)(-3)=25\text{,}$ is positive. Indeed, $Q(x)$ factors as $(x+1)(2x-3)\text{.}$ Let us put this in the form of a table:

factor	degree	multiplicity
$x+1$	$1$	$1$
$2x-3$	$1$	$1$

According to Theorem 3.37, each irreducible factor of $Q(x)$ corresponding to a group of summands. The multiplicity of that factor is the number of summands in that group (one for each power). And the degree of that factor determines the form of the numerator in each of the summand in that group. If the degree is 1 then the corresponding numerator is simply a constant. If the degree is 2, then the corresponding numerator is of a linear form.

We can find $A_{11}$ and $A_{21}$ by comparing coefficients which transfers the problem into one about solving a system of linear equations (and hence can be handled by linear algebra in general). For simplicity, we relabel the coefficients $A_{11}$ and $A_{21}$ as $A_1$ and $A_2\text{,}$ respectively.

By comparing of the numerators, we get

and so,

Solving this system of linear equations yields $A_1=2/5,A_2=1/5\text{,}$ therefore the partial fractions decomposition of $F(x)$ is

Let us compute the partial fraction decomposition of

This time $\mathrm{deg}P$ is not smaller than the degree of $\mathrm{deg}Q$ (both are of degree $5$). So, the polynomial part of the decomposition is not zero. We can find that out by dividing $P$ by $Q\text{,}$ say using long division. We find that$P(x)=2Q(x)+(x^4-x^3+4x^2-3x+3)\text{.}$ Thus,

One checks that $R(x):=x^4-x^3+4x^2-3x+3$ and $Q(x)$ have no irreducible factors in common. By uniqueness of partial fraction, it remains to compute the partial fraction decomposition of $R/Q\text{.}$ Again we organize the information obtained from the factorization of $Q(x)$ in a table:

factor	degree	multiplicity
$x-1$	$1$	$1$
$x^2+1$	$2$	$2$

Thus, the decomposition of $R(x)/Q(x)$ has the form:

By clearing the denominators, we arrive at

The identity holds for all $x$ so instead of comparing coefficients, one can also derive linear relations among the coefficients by setting $x$ to various values. We aim at getting these relations as simple as possible, for instance, by setting $x=1\text{,}$ we arrive at

That is $4=4A_{11}$ and so $A_{11}=1\text{.}$ Next by setting $x=0\text{,}$ equivalently by comparing the constant terms on both sides, we get $3=A_{11}-C_{11}-C_{12}=1-C_{11}-C_{12}\text{.}$ So, $C_{11}+C_{12}=-2\text{.}$

We can mix these techniques: at this point by comparing the leading coefficients on both sides, we get $1=A_{11}+B_{11}=1+B_{11}\text{,}$ so $B_{11}=0\text{.}$ Comparing the coefficients of the $x$ term leads us to

and so $B_{12}=1\text{.}$ At this juncture, we need one more relation among $C_{11}$ and $C_{12}\text{.}$ We can do that by substituting another value for $x\text{.}$ For example, we get $-3=2C_{11}+C_{12}$ by setting $x=-1$ (checks that yourself). And so together with $C_{11}+C_{12}=-2$ we conclude that $C_{11}=-1=C_{12}\text{.}$

Putting all these together, the partial fraction decomposition of $P/Q$ is

Let us compute the partial fraction decomposition of

Note that $Q(x)$ has degree larger than $2$ and so must be reducible over the real numbers. In fact, the intermediate value theorem guarantees a real root for a cubic polynomial over $\mathbb{R}$ and hence a linear factor. For instance, in this particular case,

so $Q(x)$ must of a root between $-2$ and $0\text{.}$ Keep in mind, sketching the graph of a function often give you some insights on where are the roots located.

In general, the roots of a cubic can be expressed in terms of its coefficients by arithmetic operations and taking surds. However, we can guess the "nice" roots if there are some. Suppose we have a factorization

where $r$ is an integer. Then $-rc=10$ and hence $r$ must be a factor of $10\text{.}$ And given we know that there is a root of $Q(x)$ between $-2$ and $0\text{,}$ it is easy to check that $r=-1$ is a root of $Q(x)\text{.}$ Thus, by dividing $Q(x)$ by $(x-(-1))=(x+1)\text{,}$ we have

Since $x^2-2x+10=(x-1{)}^2+9$ has no real roots, it is irreducible. Therefore, the partial fraction decomposition of $P/Q$ looks like

By multiplying $Q(x)$ on both sides, we get to

Setting $x=-1$ yields $-13=13A\text{,}$ so $A=-1\text{.}$ Then by setting $x=0\text{,}$ we get

Thus, $C=2\text{.}$ And by comparing the coefficients of $x^2$ on both sides, we see that $2=A+B=-1+B\text{,}$ so $B=3\text{.}$

Putting all these together, we get

Let us find the integral ${\displaystyle \int \frac{x+8}{x^2+6x+34}dx\text{.}}$ Completing square we have, $x^2+6x+34=(x+3{)}^2+25\text{.}$ So the substitution $u=x+3$ turns to integral to

Using the substitution $u=5\tan \theta \text{,}$ we find

Letting $w=u^2+25\text{,}$ we get

Putting these together, we conclude that