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, csc^{2} (4x + 1).
This is a composition of two functions:

The outer function f is the csc^{2} (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) = csc^{2} u and g(x) = 4x + 1 are composed by the equality u = 4x + 1 to produce the function f(g(x)) = csc^{2} (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:

Declare a variable u and substitute it into the integral:

Differentiate u = 4x + 1 and isolate the x term.
This gives you the differential, du = 4dx.

Substitute du/4 for dx in the integral:

Evaluate the integral:

Substitute back 4x + 1 for u:
Here’s one more example. Suppose that you want to evaluate the following integral:
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
are composed by the equality u = x – 3 to produce the function
The criteria are met, so you can integrate by using the equality u = x – 3:

Declare a variable u and substitute it into the integral:

Differentiate u = x – 3 and isolate the x term.
This gives you the differential du = dx.

Substitute du for dx in the integral:

Evaluate the integral:
= ln u + C

Substitute back x – 3 for u:
= ln x – 3 + C