Question

The value of $\int \frac{1}{\left(2x-1\right)\sqrt{x^2-x}}dx$ is equal to (where $c$ is the constant of integration)

• A. $se{c}^{-1}\left(x-1\right)+c$
• B. $se{c}^{-1}\left(2x-1\right)+c$
• C. $ta{n}^{-1}x+c$
• D. $ta{n}^{-1}\left(2x-1\right)+c$

Answer 1

Answer: (B)

Let $I=\int \frac{1}{\left(2x-1\right)\sqrt{x^2-x}}dx$
Multiply denominator and numerator by $2,$ we get $I=\int \frac{2}{\left(2x-1\right)\sqrt{4x^2-4x}}dx$
$=2\int \frac{dx}{\left(2x-1\right)\sqrt{{\left(2x-1\right)}^2-1}}$
Put $2x-1=t\Rightarrow 2dx=dt$
Now, $I=\int \frac{dt}{t\sqrt{t^2-1}}$
$=se{c}^{-1}t+c$
$=se{c}^{-1}\left(2x-1\right)+c$