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

$$\int \frac{{({\mathit{x}}^4+{\mathit{x}}^7)}^{\frac{1}{4}}}{{\mathit{x}}^2}\mathit{d}\mathit{x}$$

Answered by Sutrisno last updated on 04/Dec/23

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

Commented by Calculusboy last updated on 07/Dec/23

$$\mathit{W}\mathit{o}\mathit{n}\mathit{d}\mathit{e}\mathit{r}\mathit{f}\mathit{u}\mathit{l}\mathit{s}\mathit{o}\mathit{l}\mathit{u}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathit{s}\mathit{i}\mathit{r}$$

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