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$$\begin{align} |R_5|&\leq \int_5^\infty\frac{2x+6}{(x+2)^3}dx\\ &=\int_5^\infty\left(\frac{2}{(x+2)^2}+\frac{2}{(x+2)^3}\right)dx\text{(How did we get to this step?)}\\ &=\lim_{b\rightarrow\infty}\left(-\frac{2}{(x+2)}-\frac{1}{(x+2)^2}\right)\Bigg\vert_5^b\text{(How did we change the signs and lose a power?)}\\ &=\lim_{b\rightarrow\infty}\left(\frac{2}{7}+\frac{1}{49}-\frac{2}{b+2}-\frac{1}{(b+2)^2}\right)\text{(How do we just lose b in the next step?)}\\ &=\frac{2}{7}+\frac{1}{49}\\ &=\frac{15}{49}\\ &=0.3061... \end{align}$$ How do you get to the point where math starts to click and you enjoy it?

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Answer to very last question: I don't think anyone enjoys doing routine integrals... –  wj32 Nov 7 '12 at 8:28
    
For the first step you could look at partial fractions (useful generally for integrals) - or in this simple case note that $2x+6=2(2x+2)+2$. For changing signs and "losing" a power, you have also lost the integral sign - can you integrate $\frac 1 {x^n}$? –  Mark Bennet Nov 7 '12 at 8:36

2 Answers 2

up vote 5 down vote accepted

The first step is just algebra:

$$\frac{2x+6}{(x+2)^3}=\frac{2(x+2)+2}{(x+2)^3}=\frac{2(x+2)}{(x+2)^3}+\frac2{(x+2)^3}=\frac2{(x+2)^2}+\frac2{(x+2)^3}\;.$$

The idea is to split off from $2x+6$ as much as can be cancelled with a factor of $x+2$ in the denominator, so that you end up with fractions having of the form $\frac{a}{(x+2)^n}$ for constant $a$. The reason for this is that $$\frac{a}{(x+2)^n}=a(x+2)^{-n}\;,$$ which is easily integrated by the power rule if $n\ne 1$, and into a natural log if $n=1$. And that’s what happens in the next step:

$$\int 2(x+2)^{-2}dx=2\int(x+2)^{-2}dx=2\frac{(x+2)^{-1}}{-1}=-\frac2{x+2}$$ and

$$\int 2(x+2)^{-3}dx=2\int(x+2)^{-3}dx=2\frac{(x+2)^{-2}}{-2}=-\frac1{(x+2)^2}\;.$$

If you have to, you can make the substitution $u=x+2$, $du=dx$, and the application of the power rule will be completely obvious.

The last step about which you had a question is just taking the limit:

$$\begin{align*} \lim_{b\to\infty}\left(\frac{2}{7}+\frac{1}{49}-\frac{2}{b+2}-\frac{1}{(b+2)^2}\right)&=\lim_{b\to\infty}\frac27+\lim_{b\to\infty}\frac1{49}-\lim_{b\to\infty}\frac2{b+2}-\lim_{b\to\infty}\frac1{(b+2)^2}\\ &=\frac{2}{7}+\frac{1}{49}-0-0\;. \end{align*}$$

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1: $$ \frac{2x+6}{(x+2)^3}=\frac{2x+4+2}{(x+2)^3}=\frac{2(x+2)}{(x+2)^3} +\frac{2}{(x+2)^3}=\frac{2}{(x+2)^2}+\frac{2}{(x+2)^3} $$

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