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Question Number 67617 by aliesam last updated on 29/Aug/19

Commented by kaivan.ahmadi last updated on 29/Aug/19

$$\int \frac{\sqrt[4]{x^2-2\sqrt[4]{x^6}+x}}{\sqrt[4]{x^3}}dx=\int (\frac{1}{\sqrt[4]{x}}-2\sqrt[4]{x^3}+\frac{1}{\sqrt[4]{x^2}})dx=$$$$\int (x^{\frac{-1}{4}}-2x^{\frac{3}{4}}+x^{\frac{-1}{2}})dx=\frac{4}{3}{x}^{\frac{3}{4}}-\frac{8}{7}{x}^{\frac{7}{4}}+2x^{\frac{1}{2}}+C=$$$$\frac{4}{3}\sqrt[4]{x^3}-\frac{8}{7}x\sqrt[4]{x^3}+2\sqrt{x}+C$$

Commented by mathmax by abdo last updated on 29/Aug/19

$$letA=\int \frac{cos\left(6x\right)}{cos\left(3x\right)-sin\left(3x\right)}dxchangement3x=tgive$$$$A=\frac{1}{3}\int \frac{cos\left(2t\right)}{cost-sint}dx=\frac{1}{3}\int \frac{cos\left(2t\right)(cost+sint)}{{cos}^{2}t-{sin}^{2}t}dt$$$$=\frac{1}{3}\int \frac{cos\left(2t\right)(cost+sint)}{cos\left(2t\right)}dt=\frac{1}{3}\int (cost+sint)dt$$$$=\frac{1}{3}(sint-cost)+C.$$

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