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If ∫1(x−1)(x−2) dx=log⁡∣(x−m)+ax2+bx+n∣+C\int \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,am,n,a and bb are constant, then 2m+n+a+b2m+n+a+b is equal to:

Solution

✅ Correct Option: 1

(x−1)(x−2)=x2−3x+2=(x−32)2−14(x-1)(x-2)=x^2-3x+2=\left(x-\frac32\right)^2-\frac14. Using ∫dxX2−a2=log⁡∣X+X2−a2∣\int\frac{dx}{\sqrt{X^2-a^2}}=\log|X+\sqrt{X^2-a^2}|, the result is log⁡∣(x−32)+x2−3x+2∣+C\log\left|\left(x-\frac32\right)+\sqrt{x^2-3x+2}\right|+C. So m=32m=\frac32, a=1a=1, b=−3b=-3, n=2n=2, giving 2m+n+a+b=3+2+1−3=32m+n+a+b=3+2+1-3=3.

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