Question-204233

Question Number 204233 by Perelman last updated on 09/Feb/24

Answered by Frix last updated on 09/Feb/24

∫(x^(1/2) /(x^(3/4) +1))dx =^(t=x^(1/4) ) 4∫(t^5 /(t^3 +1))dt= =4∫t^2 dt−(4/3)∫(dt/(t+1))−(4/3)∫((2t−1)/(t^2 −t+1))dt= =((4t^3 )/3)−((4ln (t+1))/3)−((4ln(t^2 −t+1))/3)= =(4/3)(t^3 −ln (t^3 +1))= =(4/3)(x^(3/4) −ln (x^(3/4) +1))+C

$$\int\frac{{x}^{\frac{\mathrm{1}}{\mathrm{2}}} }{{x}^{\frac{\mathrm{3}}{\mathrm{4}}} +\mathrm{1}}{dx}\:\overset{{t}={x}^{\frac{\mathrm{1}}{\mathrm{4}}} } {=}\:\mathrm{4}\int\frac{{t}^{\mathrm{5}} }{{t}^{\mathrm{3}} +\mathrm{1}}{dt}= \\ $$$$=\mathrm{4}\int{t}^{\mathrm{2}} {dt}−\frac{\mathrm{4}}{\mathrm{3}}\int\frac{{dt}}{{t}+\mathrm{1}}−\frac{\mathrm{4}}{\mathrm{3}}\int\frac{\mathrm{2}{t}−\mathrm{1}}{{t}^{\mathrm{2}} −{t}+\mathrm{1}}{dt}= \\ $$$$=\frac{\mathrm{4}{t}^{\mathrm{3}} }{\mathrm{3}}−\frac{\mathrm{4ln}\:\left({t}+\mathrm{1}\right)}{\mathrm{3}}−\frac{\mathrm{4ln}\left({t}^{\mathrm{2}} −{t}+\mathrm{1}\right)}{\mathrm{3}}= \\ $$$$=\frac{\mathrm{4}}{\mathrm{3}}\left({t}^{\mathrm{3}} −\mathrm{ln}\:\left({t}^{\mathrm{3}} +\mathrm{1}\right)\right)= \\ $$$$=\frac{\mathrm{4}}{\mathrm{3}}\left({x}^{\frac{\mathrm{3}}{\mathrm{4}}} −\mathrm{ln}\:\left({x}^{\frac{\mathrm{3}}{\mathrm{4}}} +\mathrm{1}\right)\right)+{C} \\ $$

Commented by Perelman last updated on 09/Feb/24

Thank you sir!

Commented by Frix last updated on 09/Feb/24

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Answered by Frix last updated on 09/Feb/24

Another possibility but maybe hard to see: ∫(x^(1/2) /(x^(3/4) +1))dx=∫x^(−(1/4)) dx−∫(dx/(x^(3/4) +1))×(1/x^(1/4) ) ∫x^(−(1/4)) dx=((4x^(3/4) )/3) −∫(dx/(x^(3/4) +1))×(1/x^(1/4) )=−(4/3)∫(1/(x^(3/4) +1))×d(x^(3/4) +1)= =−((4ln (x^(3/4) +1))/3) ⇒ ∫(x^(1/2) /(x^(3/4) +1))dx=(4/3)(x^(3/4) −ln (x^(3/4) +1))+C

$$\mathrm{Another}\:\mathrm{possibility}\:\mathrm{but}\:\mathrm{maybe}\:\mathrm{hard}\:\mathrm{to}\:\mathrm{see}: \\ $$$$\int\frac{{x}^{\frac{\mathrm{1}}{\mathrm{2}}} }{{x}^{\frac{\mathrm{3}}{\mathrm{4}}} +\mathrm{1}}{dx}=\int{x}^{−\frac{\mathrm{1}}{\mathrm{4}}} {dx}−\int\frac{{dx}}{{x}^{\frac{\mathrm{3}}{\mathrm{4}}} +\mathrm{1}}×\frac{\mathrm{1}}{{x}^{\frac{\mathrm{1}}{\mathrm{4}}} } \\ $$$$\int{x}^{−\frac{\mathrm{1}}{\mathrm{4}}} {dx}=\frac{\mathrm{4}{x}^{\frac{\mathrm{3}}{\mathrm{4}}} }{\mathrm{3}} \\ $$$$−\int\frac{{dx}}{{x}^{\frac{\mathrm{3}}{\mathrm{4}}} +\mathrm{1}}×\frac{\mathrm{1}}{{x}^{\frac{\mathrm{1}}{\mathrm{4}}} }=−\frac{\mathrm{4}}{\mathrm{3}}\int\frac{\mathrm{1}}{{x}^{\frac{\mathrm{3}}{\mathrm{4}}} +\mathrm{1}}×{d}\left({x}^{\frac{\mathrm{3}}{\mathrm{4}}} +\mathrm{1}\right)= \\ $$$$=−\frac{\mathrm{4ln}\:\left({x}^{\frac{\mathrm{3}}{\mathrm{4}}} +\mathrm{1}\right)}{\mathrm{3}} \\ $$$$\Rightarrow \\ $$$$\int\frac{{x}^{\frac{\mathrm{1}}{\mathrm{2}}} }{{x}^{\frac{\mathrm{3}}{\mathrm{4}}} +\mathrm{1}}{dx}=\frac{\mathrm{4}}{\mathrm{3}}\left({x}^{\frac{\mathrm{3}}{\mathrm{4}}} −\mathrm{ln}\:\left({x}^{\frac{\mathrm{3}}{\mathrm{4}}} +\mathrm{1}\right)\right)+{C} \\ $$

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