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Question Number 114395 by mathmax by abdo last updated on 18/Sep/20

$$\mathrm{find}{\int }_0^{\mathrm{\infty }}\frac{{(1+\mathrm{x})}^{-\frac{3}{4}}-{(1+\mathrm{x})}^{-\frac{1}{4}}}{\mathrm{x}}\mathrm{dx}$$

Answered by Olaf last updated on 19/Sep/20

$$\mathrm{I}={\int }_0^{\mathrm{\infty }}\frac{{(1+x)}^{-\frac{3}{4}}-{(1+x)}^{-\frac{1}{4}}}{x}dx$$$$u={(1+x)}^{\frac{1}{4}}$$$$du=\frac{1}{4}{(1+x)}^{-\frac{3}{4}}dx=\frac{dx}{4u^3}$$$$\mathrm{I}={\int }_1^{\mathrm{\infty }}\frac{\frac{1}{u^3}-\frac{1}{u}}{u^4-1}.4u^{3}du$$$$\mathrm{I}=4{\int }_1^{\mathrm{\infty }}\frac{1-u^2}{u^4-1}du$$$$\mathrm{I}=-4{\int }_1^{\mathrm{\infty }}\frac{du}{1+u^2}$$$$\mathrm{I}=-4{\left[\arctan u\right]}_1^{\mathrm{\infty }}=-4(\frac{\pi }{2}-\frac{\pi }{4})=-\pi$$$$$$$$$$$$$$

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