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Question Number 83649 by 698148290 last updated on 04/Mar/20

Answered by MJS last updated on 05/Mar/20

$$=\int \frac{\sqrt{1-x}}{x}dx+\int \frac{\sqrt{2-x^2}}{x}dx+\int \frac{\sqrt{4-x^4}}{x}dx$$$$\mathrm{substitute}$$$$u=\sqrt{1-x}\to dx=-2\sqrt{1-x}du$$$$v=\sqrt{2-x^2}\to dx=-\frac{\sqrt{2-x^2}}{x}dv$$$$w=\sqrt{4-x^4}\to dx=-\frac{\sqrt{4-x^4}}{2x^3}dw$$$$\mathrm{and}\mathrm{the}\mathrm{rest}\mathrm{is}\mathrm{easy}$$

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