The road to wisdom?

Well, it's plain

and simple to express.

Err and err and err again,

but less and less and less.

Calculus 2, Fall 2016

The road to wisdom?

Well, it's plain

and simple to express.

Err and err and err again,

but less and less and less.

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.

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

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

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

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

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

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

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

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

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.

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

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

\(x^4+1\) 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.

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.

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.

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

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 \[\pm \bigg\lbrace \frac{ \mbox{divisors of } a_0 }{ \mbox{divisors of } a_n } \bigg\rbrace.\]

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