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Question Number 83960 by john santu last updated on 08/Mar/20

$$\int \frac{\mathrm{x}-1}{\sqrt{{\mathrm{x}}^2-\mathrm{x}}}\mathrm{dx}?$$

Commented by mathmax by abdo last updated on 08/Mar/20

$$I=\int \frac{x-1}{\sqrt{x^2-x}}dx\Rightarrow I=\frac{1}{2}\int \frac{2x-1-1}{\sqrt{x^2-x}}dx$$$$=\frac{1}{2}\int \frac{2x-1}{\sqrt{x^2-x}}dx-\frac{1}{2}\int \frac{dx}{\sqrt{x^2-x}}wehave$$$$\int \frac{2x-1}{2\sqrt{x^2-x}}dx=\sqrt{x^2-x}+c_1$$$$\int \frac{dx}{\sqrt{x^2-x}}=\int \frac{dx}{\sqrt{x(x-1)}}=\int \frac{dx}{\sqrt{x}\sqrt{x-1}}{=}_{\sqrt{x}=t}\int \frac{2tdt}{t\sqrt{t^2-1}}$$$$=2\int \frac{dt}{\sqrt{t^2-1}}=2ln(t+\sqrt{t^2-1})+c_2=2ln(\sqrt{x}+\sqrt{x-1})+c_2\Rightarrow$$$$I=\sqrt{x^2-x}-ln(\sqrt{x}+\sqrt{x-1})+C$$

Answered by john santu last updated on 08/Mar/20

Answered by Kunal12588 last updated on 08/Mar/20

$$x-1=A\frac{d}{dx}(x^2-x)+B$$$$\Rightarrow x-1=2Ax-(A-B)$$$$A=\frac{1}{2},A-B=1\Rightarrow B=-\frac{1}{2}$$$$I=\frac{1}{2}\int \frac{\frac{d}{dx}(x^2-x)}{\sqrt{x^2-x}}-\frac{1}{2}\int \frac{dx}{\sqrt{{(x-\frac{1}{2})}^2-{\left(\frac{1}{2}\right)}^2}}$$$$I=\sqrt{x^2-x}-\frac{1}{2}log\mid x-\frac{1}{2}+\sqrt{x^2-x}\mid +c$$$$I=\sqrt{x^2-x}-\frac{1}{2}log\mid 2x-1+2\sqrt{x^2-x}\mid +c$$$$I=\sqrt{x^2-x}-\frac{1}{2}log\mid {\left(\sqrt{x}\right)}^2+2\sqrt{x}\sqrt{x-1}+{\left(\sqrt{x-1}\right)}^2\mid +c$$$$I=\sqrt{x^2-x}-log\mid \sqrt{x}+\sqrt{x-1}\mid +c$$

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