Single Variable Calculus

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Single Variable Calculus, Chapter 5, 5.5, Section 5.5, Problem 2

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Find the integral $\int x^3 (2 + x^4)^5 dx$, by making $u = 2 + x^4$

If $u = 2 + x^4$, then $du = 4x^3 dx$, so $\displaystyle x^3 dx = \frac{1}{4} du$. And

$ \begin{equation} \begin{aligned} \int x^3 (2 + x^4)^5 dx =& \int u ^5 \frac{du}{4} \\ \\ \int x^3 (2 + x^4)^5 dx =& \frac{1}{4} \int u^5 du \\ \\ \int x^3 (2 + x^4)^5 dx =& \frac{1}{4} \cdot \frac{u^{5 + 1}}{5 + 1} + C \\ \\ \int x^3 (2 + x^4)^5 dx =& \frac{1}{4} \cdot \frac{u^6}{6} + C \\ \\ \int x^3 (2 + x^4)^5 dx =& \frac{u^6}{24} + C \\ \\ \int x^3 (2 + x^4)^5 dx =& \frac{(2 + x^4)^6}{24} + C \end{aligned} \end{equation} $

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