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Question Number 119821 by bemath last updated on 27/Oct/20

$$\underset{-3}{\stackrel{0}{\int }}\frac{6x-6}{\sqrt{x^2-2x+1}}dx=?$$

Answered by bobhans last updated on 27/Oct/20

$$\underset{-3}{\stackrel{0}{\int }}\frac{6(x-1)}{\sqrt{{(x-1)}^2}}dx=\underset{-3}{\stackrel{0}{\int }}\frac{6(x-1)}{\mid x-1\mid }dx$$$$=-6\underset{-3}{\stackrel{0}{\int }}dx=-6\left(x\right)]{}_{-3}^0=-6(0+3)=-18$$

Answered by Olaf last updated on 27/Oct/20

$$x^2-2x+1\mathrm{is}\mathrm{defined}\mathrm{for}x\in [-3;0]$$$$\mathrm{I}={\int }_{-3}^0\frac{6(x-1)}{\sqrt{{(x^2-1)}^2}}dx$$$$\mathrm{I}={\int }_{-3}^0\frac{6(x-1)}{\mid x-1\mid }dx$$$$\mathrm{I}={\int }_{-3}^0\frac{6(x-1)}{1-x}dx=-6(0+3)=-18$$

Answered by 1549442205PVT last updated on 27/Oct/20

$$\mathrm{I}=\underset{-3}{\stackrel{0}{\int }}\frac{6x-6}{\sqrt{x^2-2x+1}}dx=$$$$={\int }_{-3}^0\frac{3\mathrm{d}({\mathrm{x}}^2-2\mathrm{x}+1)}{\sqrt{{\mathrm{x}}^2-2\mathrm{x}+1}}\mathrm{dx}=3{\int }_{16}^1\frac{\mathrm{du}}{\sqrt{\mathrm{u}}}$$$$=6\sqrt{\mathrm{u}}{\mid }_{16}^1=6-24=-18$$

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