A value of α such that ∫αα+1 dx/(x+α)(x+α+1) = loge(98) is

Correct option is D. -2;
Let I = ∫αα+1 dx/(x+α)(x+α+1)
Using partial fraction,
1/(x+α)(x+α+1) = A/(x+α) + B/(x+α+1)
1 = A(x+α+1) + B(x+α)
If x = -α, then 1 = A
If x = -α-1, then 1 = -B, so B = -1
Therefore, 1/(x+α)(x+α+1) = 1/(x+α) - 1/(x+α+1)
I = ∫αα+1 [1/(x+α) - 1/(x+α+1)]dx
= [ln|x+α| - ln|x+α+1|]αα+1
= [ln|(x+α)/(x+α+1)|]αα+1
= ln|(2α+1)/(2α+2)| - ln|1/2|
= ln|2(2α+1)/(2α+2)| = ln|(2α+1)/(α+1)|
Given that I = ln(98)
ln|(2α+1)/(α+1)| = ln(98)
(2α+1)/(α+1) = 98
2α+1 = 98α + 98
96α = -97
α = -97/96
Let's check the options.
If α = 2, (2α+1)/(α+1) = 5/3 ≠ 98
If α = -1/2, (2α+1)/(α+1) = 0, which is not possible.
If α = 1/2, (2α+1)/(α+1) = 2/(3/2) = 4/3 ≠ 98
If α = -2, (2α+1)/(α+1) = -3/-1 = 3 ≠ 98
There must be a mistake in the question or the options.