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Find the following integral\int x^3(2x+5)^{1'4} dx

Question:

Find the following integral {eq}\int x^3(2x+5)^{14} dx {/eq}

Integration by {eq}u {/eq}-substitution

The {eq}u {/eq}-substitution technique for integration is used to simplify a given integrand into a more integrable form. This simplification involves replacing an expression by another so that the resulting integral is entirely in terms of the new variable. Note that while the idea is simple, the execution and applicability of the technique is not always easy.

Answer and Explanation:

We will use the substitution {eq}u = 2x + 5 {/eq}:

$$\begin{align*} u & = 2x + 5 \Rightarrow x = \frac{u - 5}{2}\\ du &= 2 \ dx \Rightarrow dx = \frac{du}{2} \\ \\ \int x^3 (2x+5)^{14} \ dx &= \int \left ( \frac{u - 5}{2} \right )^3(u^{14})\left ( \frac{du}{2} \right ) \\ &= \frac{1}{2}\cdot \frac{1}{2^3}\int (u-5)^3 u^{14} \ du \\ &= \frac{1}{16}\int (u^3 - 15u^2 + 75u - 125)(u^{14})\ du \\ &= \frac{1}{16}\int (u^{17}-15u^{16}+75u^{15}-125u^{14})\ du \\ &= \frac{1}{16}\left ( \frac{u^{17+1}}{17+1} - \frac{15u^{16+1}}{16+1} + \frac{75u^{15+1}}{15+1} - \frac{125u^{14+1}}{14+1}\right )+C\\ &= \frac{1}{16}\left ( \frac{u^{18}}{18}-\frac{15u^{17}}{17}+ \frac{75u^{16}}{16}-\frac{125u^{15}}{15} \right ) + C\\ &= \frac{1}{16}\left ( \frac{u^{18}}{18}-\frac{15u^{17}}{17}+ \frac{75u^{16}}{16}-\frac{25u^{15}}{3} \right ) + C\\ &= \boxed{\frac{1}{16}\left ( \frac{1}{18}(2x+5)^{18}-\frac{15}{17}(2x+5)^{17} + \frac{75}{16}(2x+5)^{16}-\frac{25}{3}(2x+5)^{15} \right ) + C} &\color{blue}{\text{switch back to }x} \end{align*} $$


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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