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Question Number 53277 by 0955083339 last updated on 19/Jan/19

$$\underset{0}{\stackrel{1}{\int }}\frac{x}{{(1-x)}^{3/4}}dx=$$

Commented by maxmathsup by imad last updated on 19/Jan/19

$$letI={\int }_0^1\frac{x}{{(1-x)}^{\frac{3}{4}}}dxcha7gement1-x=tgive$$$$I={\int }_0^1\frac{1-t}{t^{\frac{3}{4}}}dt={\int }_0^1{t}^{-\frac{3}{4}}dt-{\int }_0^1{t}^{\frac{1}{4}}dt$$$$={\left[\frac{1}{-\frac{3}{4}+1}{t}^{-\frac{3}{4}+1}\right]}_0^1-{\left[\frac{1}{\frac{1}{4}+1}{t}^{\frac{1}{4}+1}\right]}_0^1=4-\frac{4}{5}=\frac{16}{5}.$$

Answered by ajfour last updated on 19/Jan/19

$$let{t}^4=1-x\Rightarrow 4t^{3}dt=-dx$$$$\underset{0}{\stackrel{1}{\int }}\frac{x}{{(1-x)}^{3/4}}dx={\int }_0^1\frac{1-t^4}{t^3}dx$$$$=4{\int }_0^1(1-t^4)dt=4-\frac{4}{5}=\frac{16}{5}.$$

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