Question Number 171108 by akolade last updated on 08/Jun/22
Answered by qaz last updated on 08/Jun/22
![∫_0 ^1 ((x^3 −2x^4 +x^5 )/((x+1)^7 ))dx=∫_1 ^2 (((x−1)^3 (x−2)^2 )/x^7 )dx =∫_1 ^2 ((x^5 −7x^4 +19x^3 −25x^2 +16x−4)/x^7 )dx =[−(1/x)+(7/(2x^2 ))−((19)/(3x^3 ))+((25)/(4x^4 ))−((16)/(5x^5 ))+(2/(3x^6 ))]_1 ^2 =(1/(960))](Q171118.png)
$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{x}^{\mathrm{3}} −\mathrm{2x}^{\mathrm{4}} +\mathrm{x}^{\mathrm{5}} }{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{7}} }\mathrm{dx}=\int_{\mathrm{1}} ^{\mathrm{2}} \frac{\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{3}} \left(\mathrm{x}−\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{x}^{\mathrm{7}} }\mathrm{dx} \\ $$$$=\int_{\mathrm{1}} ^{\mathrm{2}} \frac{\mathrm{x}^{\mathrm{5}} −\mathrm{7x}^{\mathrm{4}} +\mathrm{19x}^{\mathrm{3}} −\mathrm{25x}^{\mathrm{2}} +\mathrm{16x}−\mathrm{4}}{\mathrm{x}^{\mathrm{7}} }\mathrm{dx} \\ $$$$=\left[−\frac{\mathrm{1}}{\mathrm{x}}+\frac{\mathrm{7}}{\mathrm{2x}^{\mathrm{2}} }−\frac{\mathrm{19}}{\mathrm{3x}^{\mathrm{3}} }+\frac{\mathrm{25}}{\mathrm{4x}^{\mathrm{4}} }−\frac{\mathrm{16}}{\mathrm{5x}^{\mathrm{5}} }+\frac{\mathrm{2}}{\mathrm{3x}^{\mathrm{6}} }\right]_{\mathrm{1}} ^{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{960}} \\ $$