Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.6 Exercises - Page 555: 1

Answer

The solution is $$\int\frac{x^2}{5+x}dx=\frac{(5+x)^2}{2}-10(5+x)+25\ln| 5+x|+c.$$

Work Step by Step

We will use substitution $t=5+x$ which gives us $dx=dt$ and $x=t-5\Rightarrow x^2=(t-5)^2.$ Putting this into integral we have: $$\int\frac{x^2}{5+x}dx=\int\frac{(t-5)^2}{t}dt=\int\frac{t^2-10t+25}{t}dt=\int\frac{t^2}{t}dt-10\int\frac{t}{t}dt+25\int\frac{dt}{t}=\int tdt-10\int dt+25\int\frac{dt}{t}.$$ We know that: $\int tdt=\frac{t2}{2}+c_1,$ $\int dt=t+c_2$ and $\int\frac{dt}{t}=\ln| t| + c_3$ where $c_1,c_2,c_3$ are arbitrary constants, so using this we solve our integrals and we have: $$\int\frac{x^2}{5+x}dx=\frac{t^2}{2}-10t+25\ln |t |+c$$ where we used for $c=c_1+c_2+c_3$ which is arbitrary as well. We have to express the result in terms of $x$ and we will do that by expressing $t$ in terms of $x$: $$\int\frac{x^2}{5+x}dx=\frac{(5+x)^2}{2}-10(5+x)+25\ln|5+x|+c.$$
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