Integral Problem (2)

$\mathrm{if} \underset{-1}{\overset{\mathrm{a}}{\int }}\frac{\mathrm{x}+1}{{\left(\mathrm{x}+2\right)}^4}d\mathrm{x} = \frac{10}{81} \mathrm{and} \mathrm{a}>-2 \mathrm{determine} \mathrm{a}$

$eg. u = x+2 and du=dx$

$\underset{-1}{\overset{a}{\int }}\frac{x+2-1}{{\left(x+2\right)}^4}dx become \underset{-1}{\overset{a}{\int }}\frac{u-1}{{\left(u\right)}^4}du$

$\underset{-1}{\overset{a}{\int }}(\frac{u}{u^4}-\frac{1}{u^4})du = \underset{-1}{\overset{a}{\int }}(u^{-3}-u^{-4})du$

$\left[\frac{u^{-2}}{-2}-\frac{u^{-3}}{-3}\right] = \left[-\frac{u^{-2}}{2}+\frac{u^{-3}}{3}\right]$

$we can change u = x+2$

$\left[-\frac{{\left(x+2\right)}^{-2}}{2}+\frac{{\left(x+2\right)}^{-3}}{3}\right] and interval integral$
$\left[-\frac{{\left(a+2\right)}^{-2}}{2}+\frac{{\left(a+2\right)}^{-3}}{3}\right] - \left[-\frac{{\left(-1+2\right)}^{-2}}{2}+\frac{{\left(-1+2\right)}^{-3}}{3}\right] = \frac{10}{81}$
$\left[-\frac{{\left(a+2\right)}^{-2}}{2}+\frac{{\left(a+2\right)}^{-3}}{3}\right] - \left[-\frac{{\left(1\right)}^{-2}}{2}+\frac{{\left(1\right)}^{-3}}{3}\right] = \frac{10}{81}$
$\left[-\frac{{\left(a+2\right)}^{-2}}{2}+\frac{{\left(a+2\right)}^{-3}}{3}\right] - \left(\frac{-1}{6}\right) = \frac{10}{81}$
$\left[-\frac{{\left(a+2\right)}^{-2}}{2}+\frac{{\left(a+2\right)}^{-3}}{3}\right] = \frac{10}{81}-\frac{1}{6}$
$\left[-\frac{{\left(a+2\right)}^{-2}}{2}+\frac{{\left(a+2\right)}^{-3}}{3}\right] = \frac{-21}{486}$
$\left[\frac{{\left(a+2\right)}^{-3}}{3}-\frac{{\left(a+2\right)}^{-2}}{2}\right] = \frac{-21}{486}$
$\left[\frac{2{\left(a+2\right)}^2-3{\left(a+2\right)}^3}{{\left(a+2\right)}^5}\right] = \frac{-21}{486}$
$\left[\frac{2{\left(a+2\right)}^2-3{\left(a+2\right)}^3}{6{\left(a+2\right)}^5}\right] = \frac{-21}{486}=\frac{-63}{1458} if we input a=1$
$\left[\frac{2{\left(1+2\right)}^2-3{\left(1+2\right)}^3}{6{\left(1+2\right)}^5}\right] = \frac{-63}{1458} so a= 1$