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Question Number 201227 by Calculusboy last updated on 02/Dec/23

∫ (((x^4 +x^7 )^(1/4) )/x^2 )dx

$$\:\int\:\frac{\left(\boldsymbol{{x}}^{\mathrm{4}} +\boldsymbol{{x}}^{\mathrm{7}} \right)^{\frac{\mathrm{1}}{\mathrm{4}}} }{\boldsymbol{{x}}^{\mathrm{2}} }\boldsymbol{{dx}} \\ $$

Answered by Sutrisno last updated on 04/Dec/23

=∫ (((x^4 (1+x^3 ))^(1/4) )/x^2 )dx = ∫ ((x(1+x^3 ))^(1/4) )/x^2 )dx = ∫ (((1+(x^(3/2) )^2 )^(1/4) )/x)dx misal x^(3/2) =tanθ→(√x)=tan^(1/3) θ dx=((2sec^2 θ dθ)/(3(√x)))=((2sec^2 θ dθ)/(3tan^(1/3) θ)) = ∫ (((1+(tanθ)^2 )^(1/4) )/(tan^(2/3) θ)).((2sec^2 θ dθ)/(3tan^(1/3) θ)) = ∫ ((2(sec^2 θ)^(1/4) sec^2 θdθ)/(3tanθ)) = (2/3)∫ (((sec^(1/2) θ)sec^2 θdθ)/(tanθ)) = (2/3)∫ (((sec^(1/2) θ)(tan^2 θ+1)dθ)/(tanθ)) = (2/3)∫ sec^(1/2) θ(tanθ+(1/(tanθ)))dθ = (2/3)∫ (1/( (√(cosθ))))(((sinθ)/(cosθ))+((cosθ)/(sinθ)))dθ = (2/3)(∫ sinθcos^(−(3/2)) θ dθ +∫((√(cosθ))/(sinθ))dθ) =∫((√(cosθ))/(sinθ))dθ misal (√(cosθ))=u cosθ=u^2 → dθ=((−2u)/(sinθ))du =(2/3)(2cos^(−(1/2)) θ+∫(u/(sinθ)).((−2u)/(sinθ))du) =(2/3)((2/( (√(cosθ))))−2∫(u^2 /(1−cos^2 θ))du) =(2/3)((2/( (√(cosθ))))−2∫(u^2 /(1−u^4 ))du) =(2/3)((2/( (√(cosθ))))−2∫(u^2 /((1+u^2 )(1−u^2 )))du) =(2/3)((2/( (√(cosθ))))−(∫(1/((1+u^2 )))du+∫(1/(1−u^2 ))du)) =(2/3)((2/( (√(cosθ))))−(∫(1/((1+u^2 )))du+(1/2)∫(1/(1−u))+(1/(1+u))du)) =(2/3)((2/( (√(cosθ))))−(tan^(−1) u−(1/2)ln(1−u)+(1/2)ln(1+u)))+c =(2/3)((2/( (√(cosθ))))−tan^(−1) (√(cosθ))+(1/2)ln(1−(√(cosθ)))−(1/2)ln(1+(√(cosθ))))+c =(2/3)(2((1+x^3 ))^(1/4) −tan^(−1) (1/( ((1+x^3 ))^(1/4) ))+(1/2)ln(1−(1/( ((1+x^3 ))^(1/4) )))−(1/2)ln(1+(1/( ((1+x^3 ))^(1/4) ))))+c

$$\:=\int\:\frac{\left(\boldsymbol{{x}}^{\mathrm{4}} \left(\mathrm{1}+\boldsymbol{{x}}^{\mathrm{3}} \right)\right)^{\frac{\mathrm{1}}{\mathrm{4}}} }{\boldsymbol{{x}}^{\mathrm{2}} }\boldsymbol{{dx}} \\ $$$$=\:\int\:\frac{\left.\boldsymbol{{x}}\left(\mathrm{1}+\boldsymbol{{x}}^{\mathrm{3}} \right)\right)^{\frac{\mathrm{1}}{\mathrm{4}}} }{\boldsymbol{{x}}^{\mathrm{2}} }\boldsymbol{{dx}} \\ $$$$=\:\int\:\frac{\left(\mathrm{1}+\left(\boldsymbol{{x}}^{\frac{\mathrm{3}}{\mathrm{2}}} \right)^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{4}}} }{\boldsymbol{{x}}}\boldsymbol{{dx}} \\ $$$${misal}\:{x}^{\frac{\mathrm{3}}{\mathrm{2}}} ={tan}\theta\rightarrow\sqrt{{x}}={tan}^{\frac{\mathrm{1}}{\mathrm{3}}} \theta \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{dx}=\frac{\mathrm{2}{sec}^{\mathrm{2}} \theta\:{d}\theta}{\mathrm{3}\sqrt{{x}}}=\frac{\mathrm{2}{sec}^{\mathrm{2}} \theta\:{d}\theta}{\mathrm{3}{tan}^{\frac{\mathrm{1}}{\mathrm{3}}} \theta} \\ $$$$=\:\int\:\frac{\left(\mathrm{1}+\left({tan}\theta\right)^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{4}}} }{{tan}^{\frac{\mathrm{2}}{\mathrm{3}}} \theta}.\frac{\mathrm{2}{sec}^{\mathrm{2}} \theta\:{d}\theta}{\mathrm{3}{tan}^{\frac{\mathrm{1}}{\mathrm{3}}} \theta} \\ $$$$=\:\int\:\frac{\mathrm{2}\left({sec}^{\mathrm{2}} \theta\right)^{\frac{\mathrm{1}}{\mathrm{4}}} {sec}^{\mathrm{2}} \theta{d}\theta}{\mathrm{3}{tan}\theta} \\ $$$$=\:\frac{\mathrm{2}}{\mathrm{3}}\int\:\frac{\left({sec}^{\frac{\mathrm{1}}{\mathrm{2}}} \theta\right){sec}^{\mathrm{2}} \theta{d}\theta}{{tan}\theta} \\ $$$$=\:\frac{\mathrm{2}}{\mathrm{3}}\int\:\frac{\left({sec}^{\frac{\mathrm{1}}{\mathrm{2}}} \theta\right)\left({tan}^{\mathrm{2}} \theta+\mathrm{1}\right){d}\theta}{{tan}\theta} \\ $$$$=\:\frac{\mathrm{2}}{\mathrm{3}}\int\:{sec}^{\frac{\mathrm{1}}{\mathrm{2}}} \theta\left({tan}\theta+\frac{\mathrm{1}}{{tan}\theta}\right){d}\theta \\ $$$$=\:\frac{\mathrm{2}}{\mathrm{3}}\int\:\frac{\mathrm{1}}{\:\sqrt{{cos}\theta}}\left(\frac{{sin}\theta}{{cos}\theta}+\frac{{cos}\theta}{{sin}\theta}\right){d}\theta \\ $$$$=\:\frac{\mathrm{2}}{\mathrm{3}}\left(\int\:{sin}\theta{cos}^{−\frac{\mathrm{3}}{\mathrm{2}}} \theta\:{d}\theta\:+\int\frac{\sqrt{{cos}\theta}}{{sin}\theta}{d}\theta\right) \\ $$$$=\int\frac{\sqrt{{cos}\theta}}{{sin}\theta}{d}\theta \\ $$$${misal}\:\sqrt{{cos}\theta}={u} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{cos}\theta={u}^{\mathrm{2}} \:\rightarrow\:{d}\theta=\frac{−\mathrm{2}{u}}{{sin}\theta}{du} \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{2}{cos}^{−\frac{\mathrm{1}}{\mathrm{2}}} \theta+\int\frac{{u}}{{sin}\theta}.\frac{−\mathrm{2}{u}}{{sin}\theta}{du}\right) \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}\left(\frac{\mathrm{2}}{\:\sqrt{{cos}\theta}}−\mathrm{2}\int\frac{{u}^{\mathrm{2}} }{\mathrm{1}−{cos}^{\mathrm{2}} \theta}{du}\right) \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}\left(\frac{\mathrm{2}}{\:\sqrt{{cos}\theta}}−\mathrm{2}\int\frac{{u}^{\mathrm{2}} }{\mathrm{1}−{u}^{\mathrm{4}} }{du}\right) \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}\left(\frac{\mathrm{2}}{\:\sqrt{{cos}\theta}}−\mathrm{2}\int\frac{{u}^{\mathrm{2}} }{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left(\mathrm{1}−{u}^{\mathrm{2}} \right)}{du}\right) \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}\left(\frac{\mathrm{2}}{\:\sqrt{{cos}\theta}}−\left(\int\frac{\mathrm{1}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}{du}+\int\frac{\mathrm{1}}{\mathrm{1}−{u}^{\mathrm{2}} }{du}\right)\right) \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}\left(\frac{\mathrm{2}}{\:\sqrt{{cos}\theta}}−\left(\int\frac{\mathrm{1}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}{du}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}}{\mathrm{1}−{u}}+\frac{\mathrm{1}}{\mathrm{1}+{u}}{du}\right)\right) \\ $$$$ \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}\left(\frac{\mathrm{2}}{\:\sqrt{{cos}\theta}}−\left({tan}^{−\mathrm{1}} {u}−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}−{u}\right)+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{u}\right)\right)\right)+{c} \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}\left(\frac{\mathrm{2}}{\:\sqrt{{cos}\theta}}−{tan}^{−\mathrm{1}} \sqrt{{cos}\theta}+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}−\sqrt{{cos}\theta}\right)−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+\sqrt{{cos}\theta}\right)\right)+{c} \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{2}\sqrt[{\mathrm{4}}]{\mathrm{1}+{x}^{\mathrm{3}} }−{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\:\sqrt[{\mathrm{4}}]{\mathrm{1}+{x}^{\mathrm{3}} }}+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt[{\mathrm{4}}]{\mathrm{1}+{x}^{\mathrm{3}} }}\right)−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{4}}]{\mathrm{1}+{x}^{\mathrm{3}} }}\right)\right)+{c} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by Calculusboy last updated on 07/Dec/23

Wonderful solution sir

$$\boldsymbol{{Wonderful}}\:\boldsymbol{{solution}}\:\boldsymbol{{sir}} \\ $$

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