If \( \int\displaystyle\frac{1}{\sqrt{(x-1)(x-2)}}\,dx=\log\left|(x-m)+\sqrt{ax^2+bx+n}\right|+C \) where \( c \) is an arbitrary constant and \( m,n,a \) and \( b \) are constant, then \( 2m+n+a+b \) is equal to:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 3
Expanding the product and completing the square under the root:
\( \displaystyle (x-1)(x-2)=x^2-3x+2 \)
\( \displaystyle x^2-3x+2=\left(x-\frac{3}{2}\right)^2-\frac{1}{4} \)
By the standard integral \( \int\frac{dx}{\sqrt{u^2-k^2}}=\log\left|u+\sqrt{u^2-k^2}\right|+C \) with \( u=x-\frac{3}{2} \):
\( \displaystyle \int\frac{dx}{\sqrt{(x-1)(x-2)}}=\log\left|\left(x-\frac{3}{2}\right)+\sqrt{x^2-3x+2}\right|+C \)
Comparing with the given form:
\( \displaystyle m=\frac{3}{2},\ a=1,\ b=-3,\ n=2 \)
\( \displaystyle 2m+n+a+b=3+2+1-3=3 \)