A value of \alpha such that 

\int_{\alpha }^{\alpha +1}\frac{dx}{(x+\alpha )(x+\alpha +1)}=\log_{e}\left ( \frac{9}{8} \right ) is :

 

  • Option 1)

    -2 

     

     

  • Option 2)

    \frac{1}{2}

  • Option 3)

    -\frac{1}{2}

  • Option 4)

    2

 

Answers (1)

I= \int_{\alpha }^{\alpha +1}\frac{dx}{(x+\alpha )(x+\alpha +1)}

= \int_{\alpha }^{\alpha +1}\left ( \frac{1}{(x+\alpha )}-\frac{1}{(x+\alpha +1)} \right )dx

using partial fractions,

=\left [\ln|x+\alpha |-\ln|x+\alpha +1| \right ]^{\alpha +1}_{\alpha }

=\left [\ln\left | \frac{x+\alpha }{x+\alpha +1} \right | \right ]^{\alpha +1}_{\alpha }

=\ln\left [ \frac{\left ( 2\alpha +1 \right )^{2}}{(2\alpha +1)^{2}-1} \right ]=\ln\frac{9}{8}

\Rightarrow (2\alpha +1)^{2}=9

\Rightarrow (2\alpha +1)=\pm 3

\Rightarrow \alpha =1

&

\alpha =-2


Option 1)

-2 

 

 

Option 2)

\frac{1}{2}

Option 3)

-\frac{1}{2}

Option 4)

2

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