Skip to main content
\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)
Mathematics LibreTexts

2.1: Substitution

Recall that the chain rule states that

\[ (f(g(x)))' = f'(g(x))g'(x). \]

Integrating both sides we get:

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

or

\[ \int f'\left( g(x) \right) \, g' (x) \, dx = f\left(g(x)\right) + C \]

Example 1

Calculate

\[ \int \dfrac{2x}{x^2+1}\, dx = \int 2x\left( x^2+1\right)^{-2} \, dx. \]

Solution

Let

\[ u =  x^2 +1 \]

then

\[ \dfrac{du}{dx} = 2x \]

and

\[ du = 2x \,dx.\]

We substitute:

 \[ \int u^{-2} du = -u^{-1} + C =  (x^2 +1)^{-1}  + C. \]

Steps:

  1. Find the function derivative pair (\(f\) and \(f'\)).
  2. Let \(u = f(x)\).
  3. Find \(du/dx\) and adjust for constants.
  4. Substitute.
  5. Integrate.
  6. Resubstitute.

We will try many more examples including those such as

 \[ \int x\, \sin(x^2)\, dx, \]

\[ \int x\, \sqrt{x - 2}\, dx. \]

Contributors

  • Integrated by Justin Marshall.