The road to wisdom?

Well, it's plain
and simple to express.
Err and err and err again,
but less and less and less.

– Piet Hein

From Oakley's A Mind for Numbers (p. 110+)

Oraldo "Buddy" Saucedo's important lessons learned in his climb to success:

• You are not your grade; you are better than that. Grades are indicators of time management and a rate of success.

• Bad grades do not mean you are a bad person.

• Procrastination is the death of success.

• Focusing on taking small, manageable steps forward and time management are key.

• Preparation is key to success.

• We all have a failure rate. You will fail. So control your failures. That is why we do homework–to exhaust our failure rate.

• The biggest lie ever is that practice makes perfect. Not true–practice makes you better.

• Practice is where you're supposed to fail.

• Practice at home, in class, anytime and anywhere–except on the TEST!

• Cramming and passing are not success.

• Cramming for tests is the short game with less satisfaction and only temporary results.

Definition: A rational function is a quotient of polynomials.

1. Is it possible to find numbers \(A\) and \(B\) such that \[\frac{1}{x^2-1} = \frac{A}{x-1}+\frac{B}{x+1}?\]

2. Is it possible to find numbers \(A\) and \(B\) such that \[\frac{1}{x(x^2+1)} = \frac{A}{x} + \frac{B}{x^2+1}?\]

3. Is it possible to find numbers \(A\) and \(B\) such that \[\frac{1}{x(x-1)^2} = \frac{A}{x}+\frac{B}{x-1}?\]

1. Yes. Now integrate it. What is its domain? \[\frac{1}{x^2-1} = \frac{ 1/2 }{x-1}+\frac{ -1/2 }{x+1}.\]

2. No. Is it possible to find numbers \(A\), \(B\), and \(C\) such that \[\frac{1}{x(x^2+1)} = \frac{A}{x} + \frac{Bx+C}{x^2+1}?\]

3. No. Is it possible to find numbers \(A\), \(B\), and \(C\) such that \[\frac{1}{x(x-1)^2} = \frac{A}{x}+\frac{B}{x-1}+\frac{C}{(x-1)^2}?\]

1. On the \(x\)-intervals \((-\infty, -1)\), \((-1,1)\), or \((1,\infty)\), \[\int \frac{dx}{x^2-1} = {\textstyle\frac{1}{2}}\ln|x-1| - {\textstyle\frac{1}{2}}\ln|x+1| + C.\]

2. Yes. Integrate it. What is its domain? \[\frac{1}{x(x^2+1)} = \frac{1}{x} - \frac{x}{x^2+1}.\]

3. Yes. Now integrate it. What is its domain? \[\frac{1}{x(x-1)^2} = \frac{1}{x} - \frac{1}{x-1} + \frac{1}{(x-1)^2}.\]

Definition: A polynomial \(q(x)\) is irreducible if it cannot be factored into nonconstant polynomials. Otherwise, we say it is reducible.

Theorem: A number \(a\) is a root of a polynomial \(q(x)\) if and only if \((x-a)\) is a factor of \(q(x)\).

Theorem: A degree 2 polynomial \(q(x)\) is irreducible if and only if it has no real roots.

1. \(q(x) = x^2 + 3x + 3\) is ____________.

2. Any degree 3 polynomial is _____________ since its graph must cross the \(x\)-axis.

3. \(x^4+1\) is ______________, even though it has no real roots, since \(x^4+1 = (x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x+1)\).

When a rational function \(p(x)/q(x)\) has an irreducible quadratic factor of its denominator \(q(x)\), the corresponding numerator in a partial fraction decomposition is a linear polynomial.

Example

Find the form of the partial fraction decomposition of \[\int \frac{x^3+x^2+2x+3}{x^4+5x^2+6} \, dx.\] Evaluation trick: factor the numerator and look for cancellation.

When a rational function \(p(x) / q(x)\) has repeated irreducible factors in its denominator, each power of that irreducible factor is represented once in a partial fraction decomposition.

Example

Find the form of the partial fraction decomposition of \[\int \frac{2x+3}{x(x-1)(x+2)^3 (4x^2+5) (6x^2+7x+8)^2} \, dx.\]

When the degree of the denominator \(q(x)\) of a rational function is less than the degree of the numerator \(p(x)\), use long division before doing a partial fraction decomposition.

\(\displaystyle \int \frac{ 2x^3 - 4x^2 - x - 3 }{ x^2-2x-3 } \, dx\)

Rational root theorem

If \(q(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0\) is a polynomial with integer coefficients, then the only possible rational roots of \(q(x)\) are \[\pm \bigg\lbrace \frac{ \mbox{divisors of } a_0 }{ \mbox{divisors of } a_n } \bigg\rbrace.\]

Example

Use the rational root theorem to factor the denominator: \[\int \frac{ x^2 + 4x + 1 }{ x^3 + 3x^2 - x - 3 } \, dx\]

Sometimes, a \(u\)-substitution can turn an integral of a non-rational function in \(x\) into an integral of a rational function in \(u\).

Will the non-rational function \(\displaystyle \frac{1}{x+\sqrt[3]{x}}\) and the rational function \(\displaystyle \frac{1}{x^3+x}\) have the same integral? Why or why not?

Recall that \[\begin{array}{rcl} \displaystyle \int \frac{1}{x^3+x} \, dx & = & \displaystyle \int \frac{1}{x} \, dx - \int \frac{x}{x^2+1} \, dx \\ & = & \ln|x| - {\textstyle\frac{1}{2}} \ln(x^2+1)+C. \end{array}\]