Integrate the rewritten expression

Three terms, one answer

The top curve is \(\dfrac{(x+1)^2}{x}\) for \(x > 0\). Expand the square and divide each term by \(x\). The result is \(x + 2 + \dfrac{1}{x}\), and the faint dashed pieces are its three terms.

The lower curve is a candidate \(F\), an answer still to be tested. It adds the antiderivatives of the three terms, \(\dfrac{x^2}{2}\), \(2x\) and \(\ln x\).

\(F\) must have slope equal to the height of the top curve, and the orange dashes show that slope. Drag the orange height along \(x\). The first two readouts stay equal.

Press the button. The teal bracket shows the rise \(F(3) - F(1) = 8 + \ln 3\), about 9.10, which is the area under the top curve from 1 to 3.

Any rewritten integrand

After the rewrite, the integrand is a sum of terms, and each term has an antiderivative that you know.

The sum of these antiderivatives is an antiderivative of the whole, so the lower curve runs along the orange dashes.

Move the Constant C slider. The lower curve moves up and down, but its slope does not change, so every value of C gives an antiderivative.

The teal bracket, the rise between two inputs, is the same for every C, so the constant cancels in a definite integral.

Notation

1. Integrate each term of the rewritten integrand. The integral of a sum is the sum of the integrals, and a constant factor stays in front.

   \[\begin{aligned} \int x^{k}\,dx &= \dfrac{x^{k+1}}{k+1} + C \quad (k \neq -1) \\ \int \dfrac{1}{x}\,dx &= \ln|x| + C \end{aligned}\]

2. If a term has the form \(f(ax + b)\), divide the usual antiderivative by \(a\), because the inside function \(ax + b\) has slope \(a\).

   \[\int e^{3x}\,dx = \dfrac{1}{3}e^{3x} + C\]

3. Write the logarithm with an absolute value, because \(\ln|x|\) has slope \(1/x\) for negative \(x\) too.

   \[\dfrac{d}{dx}\ln|x| = \dfrac{1}{x}\]

4. Write one constant \(C\) at the end. For a definite integral, subtract the values of \(F\) at the limits, and the constant cancels.

   \[\int_a^b f(x)\,dx = \cb{F(b) - F(a)}\]

5. To test an answer, differentiate it. The result must be the rewritten integrand, which equals the original one.

When it fails

For negative \(x\), \(\ln x\) has no value, but \(\ln|x|\) has slope \(\tfrac1x\) there too. From \(-3\) to \(-1\) the rise of \(F\) is \(\ln|{-1}|\) \(-\,\ln|{-3}|\), which is \(-\ln 3\), because the other two terms are equal at both ends. The rise is negative, so the bracket is red.

Neither curve has a value at \(x = 0\), so \(F(b) - F(a)\) means nothing when 0 lies between the limits. Keep each integral on one side of 0.

Example 1: expand, then integrate

Find \(\displaystyle\int (2x-1)^2\,dx\).

1. Expand the square into three terms. Each term is a power of \(x\).

   \[(2x-1)^2 = 4x^2 - 4x + 1\]

2. Integrate each term with the power rule, which raises the exponent by 1 and divides by the new exponent.

   \[\cy{\dfrac{4}{3}x^3 - 2x^2 + x} + C\]

3. Differentiate the answer. The result is the expanded integrand, which equals \((2x-1)^2\), so every orange dash lies along the lower curve.

   \[\begin{aligned} &\dfrac{d}{dx}\left[\dfrac{4}{3}x^3 - 2x^2 + x\right] \\ &\quad = 4x^2 - 4x + 1 \end{aligned}\]

4. The substitution \(u = 2x - 1\) gives \(\dfrac{(2x-1)^3}{6}\). The lower curve moves down by \(\tfrac16\) and still runs along every dash, because the two answers differ by a constant.

   \[\begin{aligned} \dfrac{(2x-1)^3}{6} &= \dfrac{4}{3}x^3 - 2x^2 + x \\ &\quad - \dfrac{1}{6} \end{aligned}\]

Example 2: a negative exponent

Evaluate \(\displaystyle\int_0^1 \dfrac{(e^x+1)^2}{e^x}\,dx\).

1. Expand the square. Then divide each term by \(e^x\). The faint dashed pieces are the three terms.

   \[\dfrac{e^{2x} + 2e^x + 1}{e^x} = e^x + 2 + e^{-x}\]

2. An antiderivative of \(e^x\) is \(e^x\), and an antiderivative of 2 is \(2x\).

3. A tempting wrong move: An antiderivative of \(e^{-x}\) is \(e^{-x}\).

   \[\left[e^x + 2x + \cr{e^{-x}}\right]_0^1 = e + \dfrac{1}{e}\]

   The curve is never below the line at 4, so the area from 0 to 1 is at least 4, but \(e + \tfrac1e\) is about 3.09. The red tangent is also flatter than the orange one, so this candidate has the wrong slope.

4. The inside function \(-x\) has slope \(-1\), so an antiderivative of \(e^{-x}\) is \(e^{-x}\) divided by \(-1\), which is \(-e^{-x}\).

   \[\int e^{-x}\,dx = -e^{-x}\]

5. Evaluate from 0 to 1. At \(x = 0\) the expression is \(1 + 0 - 1 = 0\), so the value is \(e + 2 - \tfrac1e\), about 4.35.

   \[\left[e^x + 2x - e^{-x}\right]_0^1 = e + 2 - \dfrac{1}{e}\]

Practice

Evaluate \(\displaystyle\int_1^2 \dfrac{3x^2+2x+4}{x^2}\,dx\).

Divide each term of the top by \(x^2\). Then integrate each term.

Questions

The top curve is the integrand \(3x^2 - \dfrac{4}{x^2}\). The lower curve is a candidate B for an antiderivative. The orange dashes show the slope that the integrand asks for.

1. At \(x = 1\), what slope does the integrand ask for?

   • \(-1\)
   • 1
   • 7
   Answer

   \(-1\)

2. The red tangent of B has slope 7 at \(x = 1\), but the orange tangent has slope \(-1\). Which step produced B?

   • The term \(-4x^{-2}\) was integrated to \(-4x^{-1}\), without dividing by the new exponent
   • The term \(3x^2\) was integrated to \(3x^3\)
   • A constant was added
   Answer

   The term \(-4x^{-2}\) was integrated to \(-4x^{-1}\), without dividing by the new exponent

3. Evaluate \(\displaystyle\int_0^1 \left(e^{3x} + 2x\right)dx\).

   • \(\dfrac{e^3 - 1}{3} + 1\)
   • \(e^3\)
   • \(3(e^3 - 1) + 1\)
   Answer

   \(\dfrac{e^3 - 1}{3} + 1\)

Summary

You integrate each term with its own formula, divide by the slope of a linear inside function, and write one constant.

This works because the integral of a sum is the sum of the integrals, and the slope of the answer equals the integrand.

You differentiate the final answer to confirm it, and you keep each integral on one side of any x where a term has no value.

This is one of four skills in topic 6.10 of Which & Why: AP Calculus.