Find the integral: \int \frac{x}{x + 16}\;dx

Question:

Find the integral:

{eq}\displaystyle \int \frac{x}{x + 16}\;dx {/eq}

Integration by substitution:

Integration by substitution method is one of the methods to solve the integral. It is also called a u-substitution. The sum rule is used to integrate. The sum rule is {eq}\displaystyle {\color{Green}{\int f(x)\pm g(x)\;dx = \int f(x)\;dx \pm \int g(x)\;dx}} {/eq}.

Answer and Explanation:

{eq}\text{Solution}: \\ \text{Given}: \\ \displaystyle \int \frac{x}{x + 16}\;dx \\ \text{Apply u-substituton}: \\ u = x + 16 \\ du = dx \\ \text{Therefore}, \\ \begin{align*} \int \frac{x}{x + 16}\;dx &= \int \frac{u - 16}{u}\;du \\ &= \int 1 - \frac{16}{u}\;du \\ &= \int 1\;du - \int \frac{16}{u}\;du & \left ( \text{Apply the sum rule}: \;{\color{Green}{\int f(x)\pm g(x)\;dx = \int f(x)\;dx \pm \int g(x)\;dx}} \right ) \\ &= u - 16\ln |u| & \left ( \text{Apply the common integral} \right ) \\ &= x + 16 - 16\ln |x + 16| & \left ( \text{Where}, \;u = x + 16 \right ) \\ &= x + 16 - 16\ln |x + 16| + C & \left ( \text{Add the constant term to the solution} \right ) \\ \end{align*} \\ \boxed{\text{The solution of the integral} \; {\color{Blue}{\displaystyle \int \frac{x}{x + 16}\;dx = x + 16 - 16\ln |x + 16| + C}}} {/eq}


Learn more about this topic:

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How to Solve Integrals Using Substitution

from Math 104: Calculus

Chapter 13 / Lesson 5
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