Question

Evaluate: $\int \frac{1}{\sqrt{(x-1)}\sqrt{(x-2)}}dx$

• A. $\ln (|2\sqrt{x^2-3x+2}+2x-3|)+c$
• B. $\ln \left(|2\sqrt{x^2-3x+2}|\right)+c$
• C. $\ln (|2\sqrt{x^2-3x+2}-2x+3|)+c$
• D. $\ln (|\sqrt{x^2-3x+2}+2x-3|)+c$

Answer 1

The correct option is A $\ln (|2\sqrt{x^2-3x+2}+2x-3|)+c$

$\int \frac{1}{\sqrt{x-1}\times \sqrt{x-2}}dx$

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

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

$=\int \frac{1}{\sqrt{x^2-3x+2}}dx$

$I=\int \frac{1}{\sqrt{(x-3/2{)}^2-(\sqrt{5}/\sqrt{2}{)}^2}}dx$

as we know

$\int \frac{dx}{\sqrt{x^2-a^2}}=log|x+\sqrt{x^2-a^2}|+c$

So, $I=log\left(|-\frac{3}{2}+\sqrt{(x-\frac{3}{2}{)}^2-(\sqrt{5}/2{)}^2}|\right)$

$=log\left(|2x-3+2\sqrt{x^2-3x+2}|\right)+C_1$

$I=log\left(|2\sqrt{x^2-3x+2}+2x-3)\right|+C_1$

where $C_1=C-log2$.