Question Number 108761 by bemath last updated on 19/Aug/20
$$\:\:\:\frac{\vdots\frac{\mathcal{B}{e}}{\mathcal{M}{ath}}\vdots}{\bigstar} \\ $$$${If}\:\int_{−\mathrm{1}} ^{\:\:{a}} \:\frac{{x}+\mathrm{1}}{\left({x}+\mathrm{2}\right)^{\mathrm{4}} }\:=\:\frac{\mathrm{10}}{\mathrm{81}}\:,\:{then}\:{the}\:{value}\:{of} \\ $$$${a}−\mathrm{2}\:{is}\:\_\_\_ \\ $$
Answered by ajfour last updated on 19/Aug/20
$${let}\:\:{I}=\int_{−\mathrm{1}} ^{\:\:{a}} \frac{{x}+\mathrm{1}}{\left({x}+\mathrm{2}\right)^{\mathrm{4}} }{dx} \\ $$$${now}\:\:{say}\:\:\:{x}+\mathrm{2}={t}\:\: \\ $$$$\Rightarrow\:\:\:\:{I}=\int_{\mathrm{1}} ^{\:\:{a}+\mathrm{2}} \left(\frac{\mathrm{1}}{{t}^{\mathrm{3}} }−\frac{\mathrm{1}}{{t}^{\mathrm{4}} }\right){dt}\:=\frac{\mathrm{10}}{\mathrm{81}} \\ $$$$\Rightarrow\:\:\:\left(\frac{\mathrm{1}}{\mathrm{3}{t}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{2}{t}^{\mathrm{2}} }\right)\mid_{\mathrm{1}} ^{{a}+\mathrm{2}} \:=\:\frac{\mathrm{10}}{\mathrm{81}} \\ $$$$\Rightarrow\:\:\frac{\mathrm{1}}{\mathrm{3}\left({a}+\mathrm{2}\right)^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{2}\left({a}+\mathrm{2}\right)^{\mathrm{2}} }=\frac{\mathrm{10}}{\mathrm{81}}+\frac{\mathrm{1}}{\mathrm{6}} \\ $$$${say}\:\:\frac{\mathrm{1}}{{a}+\mathrm{2}}={b} \\ $$$$\Rightarrow\:\:\:\:{b}^{\mathrm{3}} −\frac{\mathrm{3}}{\mathrm{2}}{b}^{\mathrm{2}} −\frac{\mathrm{10}}{\mathrm{27}}−\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0} \\ $$$$\Rightarrow\:\:\:{b}^{\mathrm{3}} −\frac{\mathrm{3}}{\mathrm{2}}{b}^{\mathrm{2}} −\frac{\mathrm{47}}{\mathrm{54}}=\mathrm{0} \\ $$$${or}\:\:\:\left({a}+\mathrm{2}\right)^{\mathrm{3}} +\frac{\mathrm{3}}{\mathrm{2}}×\frac{\mathrm{54}}{\mathrm{47}}\left({a}+\mathrm{2}\right)−\frac{\mathrm{54}}{\mathrm{47}}=\mathrm{0} \\ $$$${say}\:\:\:{a}+\mathrm{2}=\mathrm{3}{t} \\ $$$$\Rightarrow\:\:\:\mathrm{47}{t}^{\mathrm{3}} +\mathrm{9}{t}−\mathrm{2}=\mathrm{0} \\ $$$$…. \\ $$
Answered by bobhans last updated on 19/Aug/20
Commented by bemath last updated on 19/Aug/20
$${cooll} \\ $$