# How do you integrate (xdx)/(x^2+2)^3 which has the upper and lower limits, 2 and 1 ?

Apr 3, 2018

The indefinite integral is equal to $\frac{1}{48}$.

#### Explanation:

First. compute the indefinite intregral:

$\textcolor{w h i t e}{=} \int \frac{x}{{x}^{2} + 2} ^ 3 \mathrm{dx}$

Let $u = {x}^{2} + 2$, so $\mathrm{du} = 2 x \mathrm{dx}$ or $\mathrm{dx} = \frac{1}{2 x} \mathrm{du}$:

$= \int \frac{x}{u} ^ 3 \frac{1}{2 x} \mathrm{du}$

$= \int \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{x}}}}{u} ^ 3 \frac{1}{2 \textcolor{red}{\cancel{\textcolor{b l a c k}{x}}}} \mathrm{du}$

$= \frac{1}{2} \int \frac{1}{u} ^ 3 \mathrm{du}$

$= \frac{1}{2} \int {u}^{-} 3 \mathrm{du}$

$= \frac{1}{2} \cdot \frac{{u}^{- 3 + 1}}{- 3 + 1}$

$= \frac{1}{2} \cdot \frac{{u}^{- 2}}{- 2}$

$= - {u}^{-} \frac{2}{4}$

$= - \frac{1}{4 {u}^{2}}$

Plug in ${x}^{2} + 2$ back in for $u$, and don't forget to add $C$:

$= - \frac{1}{4 {\left({x}^{2} + 2\right)}^{2}} + C$

Now, plug in $x = 2$ to this and then subtract the value when you plug in $x = 1$:

$= - \frac{1}{4 {\left({2}^{2} + 2\right)}^{2}} - \left(- \frac{1}{4 {\left({1}^{2} + 2\right)}^{2}}\right)$

$= - \frac{1}{4 {\left(4 + 2\right)}^{2}} + \frac{1}{4 {\left(1 + 2\right)}^{2}}$

$= - \frac{1}{4 \cdot {6}^{2}} + \frac{1}{4 \cdot {3}^{2}}$

$= - \frac{1}{4 \cdot 36} + \frac{1}{4 \cdot 9}$

$= - \frac{1}{144} + \frac{1}{36}$

$= - \frac{1}{144} + \frac{4}{144}$

$= \frac{3}{144}$

$= \frac{1}{48}$

That's the result. Hope this helped!