Choose among the remaining methods in your course

The numerator decides

The curve is \(f(x) = \dfrac{2x}{x^2 - 1}\), for \(x > 1\). The numerator \(2x\) is the derivative of the denominator \(x^2 - 1\).

Because of this, the substitution \(u = x^2 - 1\) turns the integral into \(\int \tfrac{du}{u}\), and the result \(\ln(x^2 - 1)\) is the teal curve. The short dashes along it show the slope an antiderivative needs at each \(x\), the height of \(f\), and the curve follows them all.

Move the Numerator slider to change \(2x\) into 1. The numerator is no longer the derivative of the denominator.

The teal curve now cuts across the dashes, which turn red, and the red dashed curve on top is its slope, not \(f\).

Press Split the fraction. The method called partial fractions writes \(\tfrac{1}{x^2-1}\) as \(\tfrac12\left(\tfrac{1}{x-1} - \tfrac{1}{x+1}\right)\), two fractions from the row for \(\tfrac1x\), and the new teal curve follows the dashes. The same denominator needed two different methods.

Try the simplest method first

Press each button. Each integrand has a candidate that follows the dashes, and the method that gives it appears under the figure.

A table row or a substitution takes a few lines. Parts and partial fractions take more, and each can fail where a simpler method works.

The table comes first, perhaps after a rewrite, then substitution, and parts or partial fractions only for the shapes that need them.

Notation

1. Ask four questions in this order, and stop at the first yes. Each names a method, and the figure shows an integrand that it solves.

2. Does the integrand match a row of the table, perhaps after a rewrite? If it does, use the row.

3. Is there an inner function whose derivative is also a factor? If there is, use substitution.

4. Is the integrand a product of two different types of function, such as \(x\sin x\)? If it is, use integration by parts, with \(u\) the factor that becomes simpler when you differentiate it.

   \[\int u\,dv = uv - \int v\,du\]

5. Is the integrand a fraction with a factorable denominator, and a numerator that is not its derivative? If it is, use partial fractions, which writes it as a sum of one simpler fraction for each factor. If the numerator's degree is not below the denominator's, divide first.

   \[\begin{aligned} &\frac{5}{(x+3)(x-2)} \\ &\quad = \frac{A}{x+3} + \frac{B}{x-2} \end{aligned}\]

6. Only BC asks the last two questions.

When it fails

The integrand \(e^{x^2}\) matches no row and is not a fraction. Its inner function \(x^2\) has the derivative \(2x\), which is not a factor, and parts only returns a harder integral. The guess \(\tfrac{e^{x^2}}{2x}\), from dividing by the derivative, cuts across every dash.

Some continuous functions have antiderivatives that no formula from the table can write. A definite integral of such a function needs a numerical method.

Example 1: a product

Find \(\displaystyle\int x\ln x\,dx\), for \(x > 0\).

1. No row matches. The inner function \(\ln x\) has the derivative \(\tfrac1x\), but \(\tfrac1x\) is not a factor, so substitution does not fit. The integrand is a product of two different types, so use parts.

2. Take \(u = \ln x\), because it becomes simpler when you differentiate it. Then \(dv = x\,dx\).

   \[du = \frac{dx}{x}, \qquad v = \frac{x^2}{2}\]

3. Apply the parts formula. The new integral is a power, and the candidate follows the dashes.

   \[\begin{aligned} &\int x\ln x\,dx \\ &\quad = \frac{x^2}{2}\ln x - \int \frac{x^2}{2}\cdot\frac{dx}{x} \\ &\quad = \frac{x^2}{2}\ln x - \int \frac{x}{2}\,dx \\ &\quad = \cb{\frac{x^2}{2}\ln x - \frac{x^2}{4}} + C \end{aligned}\]

Example 2: a product that needs no parts

Find \(\displaystyle\int x^2 e^{x^3}\,dx\).

1. The integrand is a product of \(x^2\) and \(e^{x^3}\). Check the table and substitution before you try parts.

2. A tempting wrong move: A product suggests parts. Take \(u = x^2\) and \(dv = e^{x^3}\,dx\).

   \[v = \int e^{x^3}\,dx = \;?\]

   Parts needs \(v\), an antiderivative of \(e^{x^3}\). No method gives one, as with \(e^{x^2}\), so the lower panel has no candidate to draw. Parts cannot start.

3. Substitute \(u = x^3\), the inner function. Its derivative is \(3x^2\), and \(x^2\) is a factor of the integrand, so \(x^2\,dx = \tfrac13\,du\).

   \[\begin{aligned} &\int x^2 e^{x^3}\,dx \\ &\quad = \tfrac13\int e^u\,du \\ &\quad = \cb{\tfrac13 e^{x^3}} + C \end{aligned}\]

4. A product does not need parts in every case. At \(x = 1\) the slope of the answer is \(e\), the height of the curve, so its tangent is orange.

Practice

Find \(\displaystyle\int \frac{5}{x^2 + x - 6}\,dx\), for \(x > 2\).

Compare the numerator 5 with the derivative of the denominator. If they do not match, factor the denominator.

Questions

The graph shows \(f(x) = xe^{-x}\) for \(x \ge 0\). The dashes show the slope of an antiderivative, and the teal curve is a candidate for it.

1. At \(x = 0\) the curve \(f\) is 0, so an antiderivative is flat there. Which candidate is flat at \(x = 0\) and follows the dashes everywhere?

   • \(-(x+1)e^{-x}\)
   • \(-xe^{-x}\)
   • \(-e^{-x}\)
   Answer

   \(-(x+1)e^{-x}\)

2. The curve \(f\) is highest at \(x = 1\). Where does the correct antiderivative rise the fastest?

   • At \(x = 1\)
   • At \(x = 0\)
   • At \(x = 5\)
   Answer

   At \(x = 1\)

3. Which function is \(\displaystyle\int \frac{7x}{x^2 + 9}\,dx\)?

   • \(\dfrac72\ln(x^2 + 9) + C\)
   • \(7\ln(x^2 + 9) + C\)
   • \(\dfrac73\arctan\dfrac{x}{3} + C\)
   Answer

   \(\dfrac72\ln(x^2 + 9) + C\)

Summary

You ask in order whether a table row, perhaps after a rewrite, or a substitution fits, before you try parts or partial fractions.

This works because the simple methods cost less, and the shape of the integrand shows which method can succeed.

You use parts for a product of unlike types, and partial fractions for a fraction with a factorable denominator whose numerator is not its derivative.

This is one of five skills in topic 6.14 of Which & Why: AP Calculus.