Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 146929 by phally last updated on 16/Jul/21

Commented by phally last updated on 16/Jul/21

  help me?

$$\:\:\mathrm{help}\:\mathrm{me}? \\ $$

Commented by tabata last updated on 16/Jul/21

A)let:y^6 =2x+1⇒3y^5 dy=dx    (A)=∫  ((3y^5 )/(y^4 −y^3 ))dy=3∫ ((y^2 −1+1)/(y−1))dy    (A)=3∫ [(y+1)+(1/(y−1))]dy=3(y^2 /2)+3y+3 ln∣y−1∣+C    replase y by ((2x+1))^(1/6)     ∴(A)=∫ (dx/( (((2x+1)^2 ))^(1/3) −(√(2x+1))))=(3/2)((2x+1))^(1/3) +3((2x+1))^(1/6) +3 ln∣((2x+1))^(1/6) −1∣+C  ⟨M:T⟩

$$\left.{A}\right){let}:{y}^{\mathrm{6}} =\mathrm{2}{x}+\mathrm{1}\Rightarrow\mathrm{3}{y}^{\mathrm{5}} {dy}={dx} \\ $$$$ \\ $$$$\left({A}\right)=\int\:\:\frac{\mathrm{3}{y}^{\mathrm{5}} }{{y}^{\mathrm{4}} −{y}^{\mathrm{3}} }{dy}=\mathrm{3}\int\:\frac{{y}^{\mathrm{2}} −\mathrm{1}+\mathrm{1}}{{y}−\mathrm{1}}{dy} \\ $$$$ \\ $$$$\left({A}\right)=\mathrm{3}\int\:\left[\left({y}+\mathrm{1}\right)+\frac{\mathrm{1}}{{y}−\mathrm{1}}\right]{dy}=\mathrm{3}\frac{{y}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{3}{y}+\mathrm{3}\:{ln}\mid{y}−\mathrm{1}\mid+{C} \\ $$$$ \\ $$$${replase}\:{y}\:{by}\:\sqrt[{\mathrm{6}}]{\mathrm{2}{x}+\mathrm{1}} \\ $$$$ \\ $$$$\therefore\left({A}\right)=\int\:\frac{{dx}}{\:\sqrt[{\mathrm{3}}]{\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} }−\sqrt{\mathrm{2}{x}+\mathrm{1}}}=\frac{\mathrm{3}}{\mathrm{2}}\sqrt[{\mathrm{3}}]{\mathrm{2}{x}+\mathrm{1}}+\mathrm{3}\sqrt[{\mathrm{6}}]{\mathrm{2}{x}+\mathrm{1}}+\mathrm{3}\:{ln}\mid\sqrt[{\mathrm{6}}]{\mathrm{2}{x}+\mathrm{1}}−\mathrm{1}\mid+{C} \\ $$$$\langle{M}:{T}\rangle \\ $$$$ \\ $$

Commented by tabata last updated on 16/Jul/21

(B) let w^6 =1−2x⇒−3w^5 dw=dx    (B)=∫ ((−3 w^5 dw)/(w^3 +w^2 ))=−3∫ ((w^3 +1−1)/(w+1))dw    (B)=−3∫ (w^2 −w+1−(1/(w+1)))dw=−3((w^3 /3)−(w^2 /2)+w−ln∣w+1∣)+C    replase w=((1−2x))^(1/6)     ∴(B)=∫ (dx/( (√(1−2x))+((1−2x))^(1/3) ))=(3/2)((1−2x))^(1/3) −(√(1−2x))−3((1−2x))^(1/6) +3ln∣((1−2x))^(1/6) +1∣+C    ⟨M.T⟩

$$\left({B}\right)\:{let}\:{w}^{\mathrm{6}} =\mathrm{1}−\mathrm{2}{x}\Rightarrow−\mathrm{3}{w}^{\mathrm{5}} {dw}={dx} \\ $$$$ \\ $$$$\left({B}\right)=\int\:\frac{−\mathrm{3}\:{w}^{\mathrm{5}} {dw}}{{w}^{\mathrm{3}} +{w}^{\mathrm{2}} }=−\mathrm{3}\int\:\frac{{w}^{\mathrm{3}} +\mathrm{1}−\mathrm{1}}{{w}+\mathrm{1}}{dw} \\ $$$$ \\ $$$$\left({B}\right)=−\mathrm{3}\int\:\left({w}^{\mathrm{2}} −{w}+\mathrm{1}−\frac{\mathrm{1}}{{w}+\mathrm{1}}\right){dw}=−\mathrm{3}\left(\frac{{w}^{\mathrm{3}} }{\mathrm{3}}−\frac{{w}^{\mathrm{2}} }{\mathrm{2}}+{w}−{ln}\mid{w}+\mathrm{1}\mid\right)+{C} \\ $$$$ \\ $$$${replase}\:{w}=\sqrt[{\mathrm{6}}]{\mathrm{1}−\mathrm{2}{x}} \\ $$$$ \\ $$$$\therefore\left({B}\right)=\int\:\frac{{dx}}{\:\sqrt{\mathrm{1}−\mathrm{2}{x}}+\sqrt[{\mathrm{3}}]{\mathrm{1}−\mathrm{2}{x}}}=\frac{\mathrm{3}}{\mathrm{2}}\sqrt[{\mathrm{3}}]{\mathrm{1}−\mathrm{2}{x}}−\sqrt{\mathrm{1}−\mathrm{2}{x}}−\mathrm{3}\sqrt[{\mathrm{6}}]{\mathrm{1}−\mathrm{2}{x}}+\mathrm{3}{ln}\mid\sqrt[{\mathrm{6}}]{\mathrm{1}−\mathrm{2}{x}}+\mathrm{1}\mid+{C} \\ $$$$ \\ $$$$\langle{M}.{T}\rangle \\ $$

Commented by phally last updated on 16/Jul/21

thank youw

$$\mathrm{thank}\:\mathrm{youw} \\ $$

Commented by tabata last updated on 16/Jul/21

you are welcome

$${you}\:{are}\:{welcome} \\ $$

Answered by behi834171 last updated on 16/Jul/21

A.  let: 2x+1=t^6 ⇒t=(2x+1)^(1/6) ,dx=3t^5 dt  ⇒A=∫((    3t^5 dt)/(t^4 −t^3 ))=∫((  3t^2 dt)/(t−1))=∫((  3(t^2 −1)+3)/(t−1))dt=  =∫3[t+1+(1/(t−1))]dt=  =3[(1/2)t^2 +t+ln∣t−1∣+const.=  =(3/2)((2x+1))^(1/3) +3((2x+1))^(1/6) +3ln∣[((2x+1))^(1/6) −1]∣+C  .■

$${A}. \\ $$$${let}:\:\mathrm{2}\boldsymbol{{x}}+\mathrm{1}=\boldsymbol{{t}}^{\mathrm{6}} \Rightarrow\boldsymbol{{t}}=\left(\mathrm{2}\boldsymbol{{x}}+\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{6}}} ,{dx}=\mathrm{3}{t}^{\mathrm{5}} {dt} \\ $$$$\Rightarrow{A}=\int\frac{\:\:\:\:\mathrm{3}{t}^{\mathrm{5}} {dt}}{{t}^{\mathrm{4}} −{t}^{\mathrm{3}} }=\int\frac{\:\:\mathrm{3}{t}^{\mathrm{2}} {dt}}{{t}−\mathrm{1}}=\int\frac{\:\:\mathrm{3}\left({t}^{\mathrm{2}} −\mathrm{1}\right)+\mathrm{3}}{{t}−\mathrm{1}}{dt}= \\ $$$$=\int\mathrm{3}\left[{t}+\mathrm{1}+\frac{\mathrm{1}}{{t}−\mathrm{1}}\right]{dt}= \\ $$$$=\mathrm{3}\left[\frac{\mathrm{1}}{\mathrm{2}}{t}^{\mathrm{2}} +{t}+{ln}\mid{t}−\mathrm{1}\mid+{const}.=\right. \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}\sqrt[{\mathrm{3}}]{\mathrm{2}\boldsymbol{{x}}+\mathrm{1}}+\mathrm{3}\sqrt[{\mathrm{6}}]{\mathrm{2}\boldsymbol{{x}}+\mathrm{1}}+\mathrm{3}\boldsymbol{{ln}}\mid\left[\sqrt[{\mathrm{6}}]{\mathrm{2}\boldsymbol{{x}}+\mathrm{1}}−\mathrm{1}\right]\mid+\boldsymbol{{C}}\:\:.\blacksquare \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com