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

Engineering

Mathematics

Properties of Definite Integral

Question

A value of α such that $\int_{α}^{α + 1} \frac{dx}{(x + α) (x + α + 1)} = \log_{e} (\frac{9}{8})$ is :

$- \frac{1}{2}$

–2

$\frac{1}{2}$

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$\int_{}^{} \frac{dx}{(x + α) (x + α + 1)} = \log_{e} (\frac{9}{8})$

$\Rightarrow \int_{α}^{α + 1} \frac{(x + α + 1) - (x - α)}{(x + α) (x + α + 1)} dx = \log_{e} (\frac{9}{8})$

$\Rightarrow \int_{α}^{α + 1} \frac{dx}{x + α} - \int_{α}^{α + 1} \frac{dx}{x + α + 1} = \log_{e} (\frac{9}{8})$

$\Rightarrow \log_{e} {(\frac{x + α}{x + α + 1})}_{a}^{α + 1} = \log_{e} (\frac{9}{8})$

$\Rightarrow \log_{e} (\frac{2 α + 1}{2 α + 2}) (\frac{2 α}{2 α + 1}) = \log_{e} (\frac{9}{8})$

$\Rightarrow [(\frac{2 α + 1}{2 α + 2}) (\frac{2 α + 1}{2 α})] = \log_{e} \frac{9}{8}$

$\Rightarrow \frac{{(2 α + 1)}^{2}}{4 α (α + 2)} = \frac{9}{8}$

⇒ 8[4α² + 4α + 1] = 9[4α² + 4α]

⇒ 32α² + 32α + 8 = 36α² + 36α

⇒ 4α² + 4α – 8 = 0

⇒ α² + α – 2 = 0

⇒ (α + 2) (α – 1) = 0

⇒ α = 1, –2