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Question Number 154351 by ZiYangLee last updated on 17/Sep/21

$${\int }_{-1}^0\frac{x-1}{\sqrt{x^2-4x+3}}dx=?$$

Answered by ZiYangLee last updated on 17/Sep/21

$${\int }_{-1}^0\frac{x-1}{\sqrt{x^2-4x+3}}dx$$$$={\int }_{-1}^0\frac{(x-2)+1}{\sqrt{x^2-4x+3}}dx$$$$={\int }_{-1}^0\frac{\frac{1}{2}(2x-4)+1}{\sqrt{x^2-4x+3}}dx$$$$={\int }_{-1}^0\frac{(2x-4)+2}{2\sqrt{x^2-4x+3}}dx$$$$={\int }_{-1}^0\frac{2x-4}{2\sqrt{x^2-4x+3}}dx+{\int }_{-1}^0\frac{1}{\sqrt{x^2-4x+3}}dx$$$$letu=x^2-4x+3$$$$du=(2x-4)dx$$$$={\int }_8^3\frac{1}{2\sqrt{u}}du+{\int }_{-1}^0{(x^2-4x+3)}^{-\frac{1}{2}}dx$$$$=-{\int }_3^8\frac{1}{2\sqrt{u}}du+{\left[\frac{\sqrt{x^2-4x+3}}{2(2x-4)}\right]}_{-1}^0$$$$=-\frac{1}{2}{\left[2\sqrt{u}\right]}_3^8+\frac{1}{4}{\left[\frac{\sqrt{x^2-4x+3}}{x-2}\right]}_{-1}^0$$$$=-\frac{1}{2}(4\sqrt{2}-2\sqrt{3})+\frac{1}{4}(-\frac{\sqrt{3}}{2}-0)$$$$\text{You can't use 'macro parameter character \#' in math mode}$$

Commented by ZiYangLee last updated on 17/Sep/21

$$\mathrm{Does}\mathrm{anyone}\mathrm{get}\mathrm{the}\mathrm{same}\mathrm{answer}$$$$\mathrm{with}\mathrm{me}..?$$

Answered by ARUNG_Brandon_MBU last updated on 17/Sep/21

$$I={\int }_{-1}^0\frac{x-1}{\sqrt{x^2-4x+3}}dx$$$$=\frac{1}{2}{\int }_{-1}^0\frac{(2x-4)}{\sqrt{x^2-4x+3}}dx+{\int }_{-1}^0\frac{dx}{\sqrt{x^2-4x+3}}$$$$=\frac{1}{2}{\int }_{-1}^0\frac{d(x^2-4x+3)}{\sqrt{x^2-4x+3}}+{\int }_{-1}^0\frac{dx}{\sqrt{{(x-2)}^2-1}}$$$$=\frac{1}{2}\cdot \frac{2}{1}{\left[\sqrt{x^2-4x+3}\right]}_{-1}^0+{\left[\mathrm{argch}(x-2)\right]}_{-1}^0$$$$=\sqrt{3}-2\sqrt{2}+{[\ln \mid (x-2)+\sqrt{{(x-2)}^2-1}\mid ]}_{-1}^0$$$$=\sqrt{3}-2\sqrt{2}+\ln \left(\frac{\sqrt{3}-2}{-1}\right)=\sqrt{3}-2\sqrt{2}+\ln (2-\sqrt{3})$$

Answered by peter frank last updated on 18/Sep/21

$$\mathrm{Q}\left[152270\right]$$

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