int frac{d x}{(x-2) sqrt{x^2-3 x+5}} = | vedclass

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(B) Let $I = \int \frac{d x}{(x-2) \sqrt{x^2-3 x+5}}$.Substitute $x-2 = \frac{1}{t}$, then $dx = -\frac{1}{t^2} dt$ and $x = 2 + \frac{1}{t}$.Substitu

Vedclass · Accepted Answer

$\frac{-1}{\sqrt{3}} \sinh ^{-1}\left[\frac{x+4}{\sqrt{11}(x-2)}\right]+C$

Vedclass · Answer

(B) Let $I = \int \frac{d x}{(x-2) \sqrt{x^2-3 x+5}}$.Substitute $x-2 = \frac{1}{t}$, then $dx = -\frac{1}{t^2} dt$ and $x = 2 + \frac{1}{t}$.Substituting these into the integral:$I = \int \frac{-1/t^2 dt}{(1/t) \sqrt{(2+1/t)^2 - 3(2+1/t) + 5}}$$I = -\int \frac{dt/t}{\sqrt{4 + 4/t + 1/t^2 - 6 - 3/t + 5}}$$I = -\int \frac{dt}{\sqrt{1/t^2 + 1/t + 3}} = -\int \frac{dt}{\sqrt{(1 + t + 3t^2)/t^2}} = -\int \frac{dt}{\sqrt{3t^2 + t + 1}}$$I = -\frac{1}{\sqrt{3}} \int \frac{dt}{\sqrt{t^2 + t/3 + 1/3}} = -\frac{1}{\sqrt{3}} \int \frac{dt}{\sqrt{(t + 1/6)^2 + 11/36}}$Using the formula $\int \frac{du}{\sqrt{u^2 + a^2}} = \sinh^{-1}(\frac{u}{a}) + C$:$I = -\frac{1}{\sqrt{3}} \sinh^{-1}\left( \frac{t + 1/6}{\sqrt{11}/6} \right) + C = -\frac{1}{\sqrt{3}} \sinh^{-1}\left( \frac{6t + 1}{\sqrt{11}} \right) + C$Since $t = \frac{1}{x-2}$, we have:$I = -\frac{1}{\sqrt{3}} \sinh^{-1}\left( \frac{6/(x-2) + 1}{\sqrt{11}} \right) + C = -\frac{1}{\sqrt{3}} \sinh^{-1}\left( \frac{6 + x - 2}{\sqrt{11}(x-2)} \right) + C$$I = -\frac{1}{\sqrt{3}} \sinh^{-1}\left( \frac{x + 4}{\sqrt{11}(x-2)} \right) + C$.

$\int \frac{d x}{(x-2) \sqrt{x^2-3 x+5}} =$

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