Choose substitution when it fits

Strips that stretch and shrink

The curve is \(f(x) = \dfrac{6x}{(x^2+1)^2}\), and the teal region lies under it from \(x = 0\) to \(x = 1\). Thin strips cut the region, and one strip is orange.

Move the Axis slider to relabel the horizontal axis with \(u = x^2 + 1\). Each point \(x\) moves to \(u\), so a strip of width \(dx\) becomes a strip of width \(du = 2x\,dx\): it shrinks where \(2x < 1\) and stretches where \(2x > 1\).

The height of each strip changes by the opposite factor, so its area stays the same.

On the \(u\) axis, the region runs from \(u = 1\) to \(u = 2\) under the curve \(\dfrac{3}{u^2}\). An antiderivative of \(\dfrac{3}{u^2}\) is \(-\dfrac{3}{u}\), so the area is \(3 - \tfrac32 = \tfrac32\).

The substitution fits because \(x^2 + 1\) sits inside the fraction and its derivative \(2x\) is a factor of the numerator: \(6x\,dx = 3 \cdot 2x\,dx = 3\,du\).

An inner function and its derivative

Drag the orange strip, and move the Axis slider. The axis is relabelled with \(u = g(x)\), where the inner function \(g(x)\) is a function that sits inside another function.

After the slide, a strip is \(g'(x)\) times as wide as before, because \(du = g'(x)\,dx\). Its height is divided by the same factor, so its area does not change.

If \(g'(x)\) is also a factor of the integrand, that division removes it, and the curve that remains depends on \(u\) only. The integral in \(u\) is then simpler than the integral in \(x\).

A substitution fits when the integrand has an inner function and its derivative.

Notation

1. Look for an inner function \(\cy{g(x)}\). It can sit inside a root, a power, a denominator, an exponent or a trigonometric function.

   \[u = \cy{g(x)}\]

2. Check that its derivative is also a factor of the integrand. A constant multiple of the derivative is enough, because a constant moves outside the integral.

   \[du = \cy{g'(x)}\,dx\]

3. If the derivative is a factor, replace \(g(x)\) by \(u\) and \(g'(x)\,dx\) by \(du\). No \(x\) remains, so the integral is now in \(u\).

   \[\int f(g(x))\,g'(x)\,dx = \int f(u)\,du\]

4. If the derivative is not a factor, try another inner function, rewrite the integrand, or choose another method.

When it fails

The factor must be \(g'(x)\) or a constant multiple of it. Multiply the curve by an extra factor that changes with \(x\), and relabel the axis.

The strips end under the red curve, not under the dashed curve \(f(u)\), because the extra factor still contains \(x\). The integral in \(u\) is no simpler, so this substitution does not fit.

Example 1: a root with its derivative

Find \(\displaystyle\int \frac{x}{\sqrt{x^2 + 1}}\,dx\).

1. The expression \(x^2 + 1\) sits inside the root, so take it as the inner function.

   \[u = \cy{x^2 + 1}\]

2. Its derivative is \(2x\), and \(x\) is a factor of the integrand. The factor 2 is a constant, so \(x\,dx = \tfrac12\,du\).

   \[\cy{du} = 2x\,\cy{dx}\]

3. Replace \(x\,dx\) by \(\tfrac12\,du\) and \(\sqrt{x^2+1}\) by \(\sqrt{u}\). Then integrate the power of \(u\), and return to \(x\).

   \[\begin{aligned} &\int \frac{x\,dx}{\sqrt{x^2+1}} \\ &\quad = \tfrac12\int u^{-1/2}\,du \\ &\quad = \sqrt{u} + C \\ &\quad = \sqrt{x^2 + 1} + C \end{aligned}\]

4. As a check, the derivative of \(\sqrt{x^2+1}\) is \(\frac{x}{\sqrt{x^2+1}}\), the integrand.

Example 2: a factor left over

Find \(\displaystyle\int x^3\sqrt{x^2 + 1}\,dx\).

1. Try \(u = x^2 + 1\), the inner function in the root. Then \(du = 2x\,dx\). The factor \(x^3\) holds one factor \(x\), so \(x^2\) is left over.

   \[x^3\,dx = x^2 \cdot x\,dx\]

2. A tempting wrong move: The leftover \(x^2\) is a constant, so move it outside the integral.

   \[\begin{aligned} &\int x^3\sqrt{x^2+1}\,dx \\ &\quad = \frac{x^2}{2}\int \sqrt{u}\,du \end{aligned}\]

   A constant does not change when \(x\) changes, but \(x^2\) is 0 at \(x = 0\) and 2.25 at \(x = 1.5\). The red curve uses the value 2.25 everywhere, so it meets the chalk curve only at the right end.

3. Write the leftover factor in \(u\). From \(u = x^2 + 1\), \(x^2 = u - 1\), so \(x^3\,dx = x^2 \cdot x\,dx = \tfrac12(u-1)\,du\).

   \[\begin{aligned} &\int x^3\sqrt{x^2+1}\,dx \\ &\quad = \tfrac12\int (u - 1)\sqrt{u}\,du \end{aligned}\]

4. Multiply out the product, and integrate each power of \(u\).

   \[\begin{aligned} &\tfrac12\int \left(u^{3/2} - u^{1/2}\right)du \\ &\quad = \tfrac15 u^{5/2} - \tfrac13 u^{3/2} + C \end{aligned}\]

5. Return to \(x\) by replacing \(u\) with \(x^2 + 1\).

   \[\begin{aligned} &\tfrac15 (x^2+1)^{5/2} \\ &\quad - \tfrac13 (x^2+1)^{3/2} + C \end{aligned}\]

Practice

Find \(\displaystyle\int \frac{(\ln x)^3}{x}\,dx\), for \(x > 0\).

Find the function inside the cube, and check that its derivative is the other factor of the integrand.

Questions

The region under \(\dfrac{3x^2}{1+x^3}\) from \(x = 0\) to \(x = 2\) is cut into strips of width 0.1. As you answer, the figure relabels the axis with \(u = x^3\).

1. The orange strip at \(x = 1\) has width 0.1 on the \(x\) axis. About how wide is it on the \(u\) axis?

   • 0.3
   • 0.1
   • 0.03
   Answer

   0.3

2. Where on the \(x\) axis are the strips stretched the most?

   • Near \(x = 2\)
   • Near \(x = 0\)
   • Near \(x = 1\)
   Answer

   Near \(x = 2\)

3. Which function is \(\displaystyle\int 4x(1 + x^2)^3\,dx\)?

   • \(\dfrac{(1 + x^2)^4}{2} + C\)
   • \(\dfrac{(1 + x^2)^4}{4} + C\)
   • \(\dfrac{2x^2(1 + x^2)^4}{4} + C\)
   Answer

   \(\dfrac{(1 + x^2)^4}{2} + C\)

Summary

You choose substitution when the integrand has an inner function and, up to a constant, its derivative as a factor.

It works because that derivative supplies the factor \(du\), so every \(x\) disappears and the integral depends on \(u\) only.

If the derivative is not a factor, the new integral still contains \(x\), and you choose another method.

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