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:

JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Definite Integrals

A value of

α

such that

\int_{α}^{α + 1} \frac{d x}{( x + α ) ( x + α + 1 )} = lo g_{e} (\frac{9}{8})

is:

Answer: (B)

Split the integrand by partial fractions,

\frac{1}{( x + α ) ( x + α + 1 )} = \frac{1}{x + α} - \frac{1}{x + α + 1}

. Integrating from

α

α + 1

gives

ln \frac{( 2 α + 1 ) ^{2}}{( 2 α ) ( 2 α + 2 )}

, set equal to

ln \frac{9}{8}

. So

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

; testing

α = - 2

gives

\frac{( - 3 ) ^{2}}{4 ( - 2 ) ( - 1 )} = \frac{9}{8}

, so

α = - 2

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