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Question Number 123144 by aurpeyz last updated on 23/Nov/20

$$\int \frac{x-1}{\sqrt{6x-5-x^2}}dx$$

Answered by benjo_mathlover last updated on 23/Nov/20

$$6x-5-x^2=-5-(x^2-6x)$$$$=-5-(x^2-6x+9)+9$$$$=4-{(x-3)}^2$$$$\mathrm{\varnothing }\left(x\right)=\int \frac{x-1}{\sqrt{4-{(x-3)}^2}}dx$$$$letx-3=2\sin r;dx=2\cos rdr$$$$x-1=2\sin r+2$$$$\mathrm{\varnothing }\left(x\right)=\int \frac{2(\sin r+1)\left(2\cos rdr\right)}{\sqrt{4-4\sin {}^{2}r}}$$$$\mathrm{\varnothing }\left(x\right)=2\int (\sin r+1)dr$$$$\mathrm{\varnothing }\left(x\right)=-2\cos r+2r+c$$$$\mathrm{\varnothing }\left(x\right)=-2\left(\frac{\sqrt{4-{(x-3)}^2}}{2}\right)+{2\sin }^{-1}\left(\frac{x-3}{2}\right)+c$$$$\mathrm{\varnothing }\left(x\right)=-\sqrt{6x-5-x^2}+{2\sin }^{-1}\left(\frac{x-3}{2}\right)+c$$

Answered by Dwaipayan Shikari last updated on 23/Nov/20

$$\int \frac{x-3}{\sqrt{6x-5-x^2}}+\frac{2}{\sqrt{-(x^2-6x+9)+4}}dx$$$$=-\frac{1}{2}\int \frac{6-2x}{\sqrt{6x-5-x^2}}+2\int \frac{1}{\sqrt{4-{(x-3)}^2}}dx$$$$=-\sqrt{6x-5-x^2}+2{sin}^{-1}\frac{x-3}{2}+C$$

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