A value of α such that ∫ α α + 1 d x ( x + α ) ( x + α + 1 ) = l o g e 9 8 is

Step-by-step solution

I = ∫ α α + 1 d x ( x + α ) ( x + α + 1 ) = ∫ α α + 1 1 x + α   – 1 x + α + 1   d x Using partial fractions = l n x + α - l n x + α + 1 α α + 1 = l n x + α x + α + 1   α α + 1 = l n 2 α + 1 2 α + 2 - l n 2 α 2 α + 1 = l n 2 α + 1 2 2 α + 1 2 - 1 = ln ⁡ 9 8 ⇒ ( 2 α + 1 ) 2 = 9 ⇒ 2 α + 1 = ± 3 ⇒ α = 1 or - 2 .