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

Answered by Frix last updated on 09/Feb/24

$$\int \frac{x^{\frac{1}{2}}}{x^{\frac{3}{4}}+1}dx\stackrel{t=x^{\frac{1}{4}}}{=}4\int \frac{t^5}{t^3+1}dt=$$$$=4\int t^{2}dt-\frac{4}{3}\int \frac{dt}{t+1}-\frac{4}{3}\int \frac{2t-1}{t^2-t+1}dt=$$$$=\frac{4t^3}{3}-\frac{4\ln (t+1)}{3}-\frac{4\ln (t^2-t+1)}{3}=$$$$=\frac{4}{3}(t^3-\ln (t^3+1))=$$$$=\frac{4}{3}(x^{\frac{3}{4}}-\ln (x^{\frac{3}{4}}+1))+C$$

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

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

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