Question Number 43342 by pieroo last updated on 10/Sep/18
$$\mathrm{using}\:\mathrm{the}\:\mathrm{substitution}\:\mathrm{u}=\mathrm{x}+\mathrm{2},\:\mathrm{evaluate}\:\int_{\mathrm{1}} ^{\mathrm{2}} \frac{\mathrm{x}−\mathrm{1}}{\left(\mathrm{x}+\mathrm{2}\right)^{\mathrm{4}} } \\ $$
Answered by $@ty@m last updated on 10/Sep/18
![∫_3 ^4 (((u−2)−1)/u^4 )du =∫_3 ^4 ((u−3)/u^4 )du =∫_3 ^4 ((1/u^3 )−(3/u^4 ))du =[(1/(2u^2 ))−(3/(3u^3 ))]_3 ^4 =((1/(32))−(1/(64)))−((1/(18))−(1/(27))) =(1/(64))−(1/(54)) =((−5)/(1728))](https://www.tinkutara.com/question/Q43347.png)
$$\underset{\mathrm{3}} {\overset{\mathrm{4}} {\int}}\frac{\left({u}−\mathrm{2}\right)−\mathrm{1}}{{u}^{\mathrm{4}} }{du} \\ $$$$=\underset{\mathrm{3}} {\overset{\mathrm{4}} {\int}}\frac{{u}−\mathrm{3}}{{u}^{\mathrm{4}} }{du} \\ $$$$=\underset{\mathrm{3}} {\overset{\mathrm{4}} {\int}}\left(\frac{\mathrm{1}}{{u}^{\mathrm{3}} }−\frac{\mathrm{3}}{{u}^{\mathrm{4}} }\right){du} \\ $$$$=\left[\frac{\mathrm{1}}{\mathrm{2}{u}^{\mathrm{2}} }−\frac{\mathrm{3}}{\mathrm{3}{u}^{\mathrm{3}} }\underset{\mathrm{3}} {\overset{\mathrm{4}} {\right]}} \\ $$$$=\left(\frac{\mathrm{1}}{\mathrm{32}}−\frac{\mathrm{1}}{\mathrm{64}}\right)−\left(\frac{\mathrm{1}}{\mathrm{18}}−\frac{\mathrm{1}}{\mathrm{27}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{64}}−\frac{\mathrm{1}}{\mathrm{54}} \\ $$$$=\frac{−\mathrm{5}}{\mathrm{1728}} \\ $$