Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 211400 by Spillover last updated on 08/Sep/24

            ∫_0 ^1 ((x^3 −3x^2 +3x−1)/(x^4 +4x^3 +6x^2 +4x+1))dx

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{1}}{{x}^{\mathrm{4}} +\mathrm{4}{x}^{\mathrm{3}} +\mathrm{6}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}}{dx} \\ $$$$ \\ $$

Answered by Berbere last updated on 08/Sep/24

=∫_0 ^1 (((x−1)^3 )/((x+1)^4 ))dx  =∫_0 ^1 −(1/((1+x))).(((1−x)/(1+x)))^3 dx  u=((1−x)/(1+x))⇒x=((1−u)/(1+u))⇒dx=((−2)/((1+u)^2 ))du  =−∫_0 ^1 u^3 .(du/((1+u)))=−∫_0 ^1 u^2 −u+1−(1/(1+u))du  =−(5/6)+ln(2)

$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left({x}−\mathrm{1}\right)^{\mathrm{3}} }{\left({x}+\mathrm{1}\right)^{\mathrm{4}} }{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} −\frac{\mathrm{1}}{\left(\mathrm{1}+{x}\right)}.\left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right)^{\mathrm{3}} {dx} \\ $$$${u}=\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\Rightarrow{x}=\frac{\mathrm{1}−{u}}{\mathrm{1}+{u}}\Rightarrow{dx}=\frac{−\mathrm{2}}{\left(\mathrm{1}+{u}\right)^{\mathrm{2}} }{du} \\ $$$$=−\int_{\mathrm{0}} ^{\mathrm{1}} {u}^{\mathrm{3}} .\frac{{du}}{\left(\mathrm{1}+{u}\right)}=−\int_{\mathrm{0}} ^{\mathrm{1}} {u}^{\mathrm{2}} −{u}+\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+{u}}{du} \\ $$$$=−\frac{\mathrm{5}}{\mathrm{6}}+{ln}\left(\mathrm{2}\right) \\ $$

Answered by Frix last updated on 08/Sep/24

∫_0 ^1  (((x−1)^3 )/((x+1)^4 ))dx =^(t=x+1)   =∫_1 ^2 ((1/t)−(6/t^2 )+((12)/t^3 )−(8/t^4 ))dt=  =[ln t +(6/t)−(6/t^2 )+(8/(3t^3 ))]_1 ^2 =−(5/6)+ln 2

$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\frac{\left({x}−\mathrm{1}\right)^{\mathrm{3}} }{\left({x}+\mathrm{1}\right)^{\mathrm{4}} }{dx}\:\overset{{t}={x}+\mathrm{1}} {=} \\ $$$$=\underset{\mathrm{1}} {\overset{\mathrm{2}} {\int}}\left(\frac{\mathrm{1}}{{t}}−\frac{\mathrm{6}}{{t}^{\mathrm{2}} }+\frac{\mathrm{12}}{{t}^{\mathrm{3}} }−\frac{\mathrm{8}}{{t}^{\mathrm{4}} }\right){dt}= \\ $$$$=\left[\mathrm{ln}\:{t}\:+\frac{\mathrm{6}}{{t}}−\frac{\mathrm{6}}{{t}^{\mathrm{2}} }+\frac{\mathrm{8}}{\mathrm{3}{t}^{\mathrm{3}} }\right]_{\mathrm{1}} ^{\mathrm{2}} =−\frac{\mathrm{5}}{\mathrm{6}}+\mathrm{ln}\:\mathrm{2} \\ $$

Answered by MathematicalUser2357 last updated on 10/Sep/24

−0.140186

$$−\mathrm{0}.\mathrm{140186} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com