## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 8 - Section 8.1 - Integration by Parts - Exercises - Page 427: 10

#### Answer

$$\int (x^2-2x+1)e^{2x}dx=\frac{e^{2x}}{4}(2x^2-6x+5)+C$$

#### Work Step by Step

$$A=\int (x^2-2x+1)e^{2x}dx$$ Set $u=x^2-2x+1$ and $dv=e^{2x}dx$ Then we would have $du=(2x-2)dx=2(x-1)dx$ and $v=\frac{1}{2}e^{2x}$ Using the formula $\int udv= uv-\int vdu$: $$A=\frac{1}{2}e^{2x}(x^2-2x+1)-\int\Big(\frac{1}{2}e^{2x}\times2(x-1)dx\Big)$$ $$A=\frac{e^{2x}(x-1)^2}{2}-\int(x-1)e^{2x}dx$$ Set $u=x-1$ and $dv=e^{2x}dx$ Then we would have $du=dx$ and $v=\frac{1}{2}e^{2x}$ Using the formula $\int udv= uv-\int vdu$: $$A=\frac{e^{2x}(x-1)^2}{2}-\Big(\frac{1}{2}e^{2x}(x-1)-\frac{1}{2}\int e^{2x}dx\Big)$$ $$A=\frac{e^{2x}(x^2-2x+1)}{2}-\Big(\frac{e^{2x}(x-1)}{2}-\frac{e^{2x}}{4}\Big)+C$$ $$A=\frac{e^{2x}(x^2-2x+1)}{2}-\frac{e^{2x}(x-1)}{2}+\frac{e^{2x}}{4}+C$$ $$A=\frac{e^{2x}}{4}(2x^2-4x+2-2x+2+1)+C$$ $$A=\frac{e^{2x}}{4}(2x^2-6x+5)+C$$

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