# 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*)*ax*or*ax*+*b**.*

Here’s an example. Suppose that you want to integrate the function, csc^{2} (4*x* + 1).

This is a composition of two functions:

The outer function

*f*is the csc^{2}(*u*) function.The inner function is

*g*(*x*) = 4*x*+ 1, which differentiates to the constant 4.

The composition is held together by the equality *u* = 4*x* + 1. That is, the two basic functions *f*(*u*) = csc^{2} *u* and *g*(*x*) = 4*x* + 1* *are composed by the equality *u* = 4*x* + 1 to produce the function *f*(*g*(*x*)) = csc^{2} (4*x* + 1).

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

Declare a variable

*u*and substitute it into the integral:Differentiate

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

*du*= 4*dx.*Substitute

*du*/4 for*dx*in the integral:Evaluate the integral:

Substitute back 4

*x*+ 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*