Question

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

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

Answer 1

Answer: (B)

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