Consider the function f(x) = \frac{4x + 4}{2x^2+4} and the function ...


Consider the function

{eq}f(x) = \frac{4x + 4}{2x^2+4} {/eq}

and the function

{eq}g(x)=\frac{2x+2}{x^2+2x} {/eq}.

Clearly, {eq}f(x) = g(x) {/eq} for any value of {eq}x {/eq} and thus {eq}f {/eq} and {eq}g {/eq} should be thought of as merely different notation for the same function.

Now find the integral of {eq}\int f(x)\text{d}x {/eq} and the integral of {eq}\int g(x)\text{d}x {/eq} (without factoring out 2 in the case of {eq}f {/eq}). Explain why it is not a problem that you get different solutions.

Substitution Method:

The integration method using the variable change is also known as the substitution method. This method is based on the chain rule for derivatives of a compound function. The general formula of this method can be defined as follows:

{eq}u=g(x)\\ du = g'(x) dx\\ \int \, f(x) dx = \int \, f(g(u)) g'(u) du {/eq}

Answer and Explanation:

Part 1.

{eq}\displaystyle f(x)=\frac{4x+4}{2x^2+4x}\\ \displaystyle f(x)=\frac{2(2x+2)}{2(x^2+2x)}\\ \displaystyle f(x)=\frac{2x+2}{x^2+2x}\\ \displaystyle g(x)=\frac{2x+2}{x^2+2x}\\ \displaystyle f(x)=g(x)\\ {/eq}

Part 2. Apply Variable Change

{eq}\displaystyle I_f= \int \, f(x) \, dx\\ \displaystyle I_f= \int \, \frac{4x + 4}{2x^2+4x} \, dx\\ \displaystyle u=2x^2+4x \,\,\,\,\,\,\,\,\,\,\,\, du=4x+4\\ \displaystyle I_f= \int \, \frac{1}{u} \, du\\ \displaystyle I_f= \ln {u} +C_1\\ \displaystyle I_f= \ln (2x^2+4x) +C_1\\ {/eq}

Part 3. Apply Variable Change

{eq}\displaystyle I_g= \int \, g(x) \, dx\\ \displaystyle I_g= \int \, \frac{2x+2}{x^2+2x} \, dx\\ \displaystyle u=x^2+2x \,\,\,\,\,\,\,\,\,\,\,\, du=2x+2\\ \displaystyle I_g= \int \, \frac{1}{u} \, du\\ \displaystyle I_g= \ln {u} +C_2\\ \displaystyle I_g= \ln (x^2+2x) +C_2\\ {/eq}

Part 4.

{eq}\displaystyle I_f= \ln (2x^2+4x) +C_1\\ \displaystyle I_f= \ln (2(x^2+2x)) +C_1\\ \displaystyle I_f= \ln (x^2+2x) +\ln 2 +C_1 \,\, \Longrightarrow \,\, \textrm {by applying the sum of logarithms} \\ \displaystyle I_f= \ln (x^2+2x) +\ln 2 +C_1 \,\, \Longrightarrow \,\, \textrm {by applying the sum of logarithms} \\ \displaystyle I_g= \ln (x^2+2x) +C_2 {/eq}

The solution of the indefinite integral is a family of functions that can vary in a constant.

Learn more about this topic:

How to Solve a System of Equations by Substitution

from 6th-8th Grade Math: Practice & Review

Chapter 29 / Lesson 4

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