A value of $ \alpha $ such that $ \int_{\alpha}^{\alpha+1} \frac{d x}{(x+\alpha)(x+\alpha+1)}=\log _{e}\left(\frac{9}{8}\right) $ is

Question

A value of $ \alpha $ such that $ \int_{\alpha}^{\alpha+1} \frac{d x}{(x+\alpha)(x+\alpha+1)}=\log _{e}\left(\frac{9}{8}\right) $ is, $ -\frac{1}{2} $, $ \frac{1}{2} $, $ -2 $, $ 2 $

MARKS App · Accepted Answer

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 .

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