How to Integrate Compositions of Functions

Compositions of functions — that is, one function nested inside another — are of the form f(g(x)). You can integrate them by substituting u = g(x) when

  • You know how to integrate the outer function f.

  • The inner function g(x) differentiates to a constant — that is, it’s of the form ax or ax + b.

Here’s an example. Suppose that you want to integrate the function, csc2 (4x + 1).

This is a composition of two functions:

  • The outer function f is the csc2 (u) function.

  • The inner function is g(x) = 4x + 1, which differentiates to the constant 4.

The composition is held together by the equality u = 4x + 1. That is, the two basic functions f(u) = csc2 u and g(x) = 4x + 1 are composed by the equality u = 4x + 1 to produce the function f(g(x)) = csc2 (4x + 1).

Both criteria are met, so this integral is a prime candidate for substitution using u = 4x + 1. Here’s how you do it:

  1. Declare a variable u and substitute it into the integral:

    image0.png
  2. Differentiate u = 4x + 1 and isolate the x term.

    This gives you the differential, du = 4dx.

    image1.png
  3. Substitute du/4 for dx in the integral:

    image2.png
  4. Evaluate the integral:

    image3.png
  5. Substitute back 4x + 1 for u:

    image4.png

Here’s one more example. Suppose that you want to evaluate the following integral:

image5.png

This is a composition of two functions:

  • The outer function f is a fraction — technically, an exponent of –1 — which you know how to integrate.

  • The inner function is g(x) = x – 3, which differentiates to 1.

The composition is held together by the equality u = x – 3. That is, the two basic functions

image6.png

are composed by the equality u = x – 3 to produce the function

image7.png

The criteria are met, so you can integrate by using the equality u = x – 3:

  1. Declare a variable u and substitute it into the integral:

    image8.png
  2. Differentiate u = x – 3 and isolate the x term.

    This gives you the differential du = dx.

  3. Substitute du for dx in the integral:

    image9.png
  4. Evaluate the integral:

    = ln |u| + C

  5. Substitute back x – 3 for u:

    = ln |x – 3| + C

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