Question Number 200159 by Calculusboy last updated on 15/Nov/23
Commented by 0670322918 last updated on 15/Nov/23
![∫^1 _0 ((x^3 −3x^2 +3x−1)/(x^4 +4x^3 +6x^2 +4x+1))dx= ∫_0 ^1 (((x−1)^3 )/((x+1)^4 ))dx=∫_0 ^1 (((x+1−2)^3 )/((x+1)^4 ))dx= ∫_0 ^1 (((x+1)^3 −6(x+1)^2 +12(x+1)−8)/((x+1)^4 ))dx ∫_0 ^1 ((1/(x+1))−(6/((x+1)^2 ))+((12)/((x+1)^3 ))−(8/((x+1)^4 )))dx= [ln(x+1)+(6/((x+1)))−(6/((x+1)^2 ))+(8/(3(x+1)^3 ))]_0 ^1 = ln(2)+3−(3/2)+(1/3)−6+6−(8/3)= ln(2)+(3/2)−(7/3)=((ln(64)−5)/6) ∫_0 ^1 ((x^3 −3x^2 +3x−1)/(x^4 +4x^3 +6x^2 +4x+1))dx=((ln(64)−5)/6)](https://www.tinkutara.com/question/Q200162.png)
$$\underset{\mathrm{0}} {\int}^{\mathrm{1}} \frac{{x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{1}}{{x}^{\mathrm{4}} +\mathrm{4}{x}^{\mathrm{3}} +\mathrm{6}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}}{dx}= \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\left({x}−\mathrm{1}\right)^{\mathrm{3}} }{\left({x}+\mathrm{1}\right)^{\mathrm{4}} }{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left({x}+\mathrm{1}−\mathrm{2}\right)^{\mathrm{3}} }{\left({x}+\mathrm{1}\right)^{\mathrm{4}} }{dx}= \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\left({x}+\mathrm{1}\right)^{\mathrm{3}} −\mathrm{6}\left({x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{12}\left({x}+\mathrm{1}\right)−\mathrm{8}}{\left({x}+\mathrm{1}\right)^{\mathrm{4}} }{dx} \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\left(\frac{\mathrm{1}}{{x}+\mathrm{1}}−\frac{\mathrm{6}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{12}}{\left({x}+\mathrm{1}\right)^{\mathrm{3}} }−\frac{\mathrm{8}}{\left({x}+\mathrm{1}\right)^{\mathrm{4}} }\right){dx}= \\ $$$$\left[{ln}\left({x}+\mathrm{1}\right)+\frac{\mathrm{6}}{\left({x}+\mathrm{1}\right)}−\frac{\mathrm{6}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{8}}{\mathrm{3}\left({x}+\mathrm{1}\right)^{\mathrm{3}} }\underset{\mathrm{0}} {\overset{\mathrm{1}} {\right]}}= \\ $$$${ln}\left(\mathrm{2}\right)+\mathrm{3}−\frac{\mathrm{3}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{6}+\mathrm{6}−\frac{\mathrm{8}}{\mathrm{3}}= \\ $$$${ln}\left(\mathrm{2}\right)+\frac{\mathrm{3}}{\mathrm{2}}−\frac{\mathrm{7}}{\mathrm{3}}=\frac{{ln}\left(\mathrm{64}\right)−\mathrm{5}}{\mathrm{6}} \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{{x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{1}}{{x}^{\mathrm{4}} +\mathrm{4}{x}^{\mathrm{3}} +\mathrm{6}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}}{dx}=\frac{{ln}\left(\mathrm{64}\right)−\mathrm{5}}{\mathrm{6}} \\ $$
Commented by Calculusboy last updated on 15/Nov/23
$$\boldsymbol{{thanks}}\:\boldsymbol{{sir}} \\ $$
Answered by Frix last updated on 15/Nov/23
![∫_0 ^1 (((x−1)^3 )/((x+1)^4 ))dx =^(t=x+1) ∫(((t−2)^3 )/t^4 )dt= =∫_1 ^2 (t^(−1) −6t^(−2) +12t^(−3) −8t^(−4) )dt= =[ln t +6t^(−1) −6t^(−2) +(8/3)t^(−3) ]_1 ^2 = =−(5/6)+ln 2](https://www.tinkutara.com/question/Q200163.png)
$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\left({x}−\mathrm{1}\right)^{\mathrm{3}} }{\left({x}+\mathrm{1}\right)^{\mathrm{4}} }{dx}\:\overset{{t}={x}+\mathrm{1}} {=}\:\int\frac{\left({t}−\mathrm{2}\right)^{\mathrm{3}} }{{t}^{\mathrm{4}} }{dt}= \\ $$$$=\underset{\mathrm{1}} {\overset{\mathrm{2}} {\int}}\left({t}^{−\mathrm{1}} −\mathrm{6}{t}^{−\mathrm{2}} +\mathrm{12}{t}^{−\mathrm{3}} −\mathrm{8}{t}^{−\mathrm{4}} \right){dt}= \\ $$$$=\left[\mathrm{ln}\:{t}\:+\mathrm{6}{t}^{−\mathrm{1}} −\mathrm{6}{t}^{−\mathrm{2}} +\frac{\mathrm{8}}{\mathrm{3}}{t}^{−\mathrm{3}} \right]_{\mathrm{1}} ^{\mathrm{2}} = \\ $$$$=−\frac{\mathrm{5}}{\mathrm{6}}+\mathrm{ln}\:\mathrm{2} \\ $$
Commented by Calculusboy last updated on 15/Nov/23
$$\boldsymbol{{thanks}}\:\boldsymbol{{sir}} \\ $$