i'm trying to solve this integral, but i feel lost.
So far,
$$\int\frac{(x-2)}{x\sqrt{x-1}\sqrt{x^{2}-x+1}}dx=\int\frac{(x-1)-1}{x\sqrt{x-1}\sqrt{x^{2}-(x-1)}}dx$$
Now, let $u = x-1$ then $du = dx$
$$\int\frac{(x-1)-1}{x\sqrt{x-1}\sqrt{x^{2}-(x-1)}}dx = \int\frac{u-1}{(u+1)\sqrt{u}\sqrt{(u+1)^{2}-u}}du$$
It is fine? What can i do next? I will appreciate any help. Thanks!
I did it (I think so). This is what i got.
$\int\frac{x-2}{x\sqrt{x-1}\sqrt{x^{2}-x+1}}dx$
= $\int$$\frac{x^{2}(\frac{1}{x}-\frac{2}{x^{2}})}{x\sqrt{x-1}\sqrt{x^{2}(1-\frac{1}{x}+\frac{1}{x2})}}dx$
= $\int\frac{\frac{1}{x}-\frac{2}{x^{2}}}{\sqrt{x^{2}(\frac{1}{x}-\frac{1}{x^{2}})}\sqrt{1-\frac{1}{x}+\frac{1}{x^{2}}}}dx$
Let $u=\sqrt{\frac{1}{x}-\frac{1}{x^{2}}}$ get $du=\frac{\frac{2}{x^{3}}-\frac{1}{x^{2}}}{2\sqrt{\frac{1}{x}-\frac{1}{x^{2}}}}dx$
then $dx=\frac{2\sqrt{\frac{1}{x}-\frac{1}{x^{2}}}}{\frac{2}{x^{3}}-\frac{1}{x^{2}}}du$
Replacing we get:
$-2\int\frac{du}{\sqrt{1-u^{2}}}$ (I omitted some details, but replacement is easy)
= $-2arcsen(u)=-2arcsen($ $\sqrt{\frac{1}{x}-\frac{1}{x^{2}}})$+ C