Question

Evaluate: $\int \frac{x-1}{\sqrt{x^2-x}}dx$.

Answer 1

$I=\int \frac{x-1}{\sqrt{x^2-x}}dx$

Write $(x-1)=\frac{1}{2}(2x-1)+(-\frac{1}{2})$

$I=\int (\frac{2x-1}{2\sqrt{x^2-x}}-\frac{1}{2\sqrt{x^2-x}})dx$

$2I=\int \frac{2x-1}{\sqrt{x^2-x}}dx-\int \frac{1}{2\sqrt{x^2-x}}dx\to I=I_1+I_2$

Substitute $u=x^2-x\to dx=\frac{1}{2x-1}du$ in $I_1$,

$I_1=\int \frac{1}{\sqrt{u}}du=2\sqrt{u}$

Undo substitution $u=x^2-x$

$I_1=2\sqrt{x^2-x}$

Now, substitute $u=2x-1\to dx=\frac{1}{2}du$ in $I_2$

$I_2=\int \frac{1}{\sqrt{u^2-1}}du=ln(\sqrt{u^2-1}-u)$

$I_2=ln(\sqrt{(2x-1{)}^2-1}+2x-1)$

$\therefore I=\sqrt{x^2-x}-\frac{ln(|\sqrt{(2x-1{)}^2-1}+2x-1)|)}{2}+C$