Question

A value of $\alpha $ such that $\int\limits_{\alpha }^{\alpha +1}{\dfrac{dx}{\left( x+\alpha \right)\left( x+\alpha +1 \right)}}={{\log }_{e}}\left( \dfrac{9}{8} \right)$ is
$\left( \text{A} \right)\text{ }\dfrac{\text{1}}{\text{2}}$
$\left( \text{B} \right)\text{ 2}$
$\left( C \right)\text{ -}\dfrac{\text{1}}{\text{2}}$
$\left( D \right)\text{ -2}$

Answer 1

Hint: In this question we have been given a definite integral which we will integrate with respect to $x$ and then we will equate both the sides of the expression to find the value of the term $\alpha $. We will integrate the term such that we get the left-hand side in the terms of a logarithm and then cancel out the logarithm.

Complete step by step solution:
We have the expression given as:
 $\Rightarrow \int\limits_{\alpha }^{\alpha +1}{\dfrac{dx}{\left( x+\alpha \right)\left( x+\alpha +1 \right)}}={{\log }_{e}}\left( \dfrac{9}{8} \right)$
We can see that the expression cannot be integrated directly; we will convert the numerator of the expression in the form of the denominator.
We can write the numerator as:
$\Rightarrow \int\limits_{\alpha }^{\alpha +1}{\dfrac{\left( x+\alpha +1 \right)-\left( x+\alpha \right)dx}{\left( x+\alpha \right)\left( x+\alpha +1 \right)}}={{\log }_{e}}\left( \dfrac{9}{8} \right)$
We can see that by substituting this in the numerator, the value of the numerator does not change.
Now on splitting the term in the numerator, we get:
\[\Rightarrow \int\limits_{\alpha }^{\alpha +1}{\left[ \dfrac{\left( x+\alpha +1 \right)}{\left( x+\alpha \right)\left( x+\alpha +1 \right)}-\dfrac{\left( x+\alpha \right)}{\left( x+\alpha \right)\left( x+\alpha +1 \right)} \right]dx}={{\log }_{e}}\left( \dfrac{9}{8} \right)\]
Now on cancelling the similar terms from the fraction, we get:
\[\Rightarrow \int\limits_{\alpha }^{\alpha +1}{\left[ \dfrac{1}{\left( x+\alpha \right)}-\dfrac{1}{\left( x+\alpha +1 \right)} \right]dx}={{\log }_{e}}\left( \dfrac{9}{8} \right)\]
Now we know that $\int{\dfrac{dx}{\left( x+a \right)}=\log \left( x+a \right)}$ therefore, on using the formula, we get:
\[\Rightarrow \left[ {{\log }_{e}}\left( x+\alpha \right)-{{\log }_{e}}\left( x+\alpha +1 \right) \right]_{\alpha }^{\alpha +1}={{\log }_{e}}\left( \dfrac{9}{8} \right)\]
Now we know the property of logarithm that ${{\log }_{e}}a-{{\log }_{e}}b={{\log }_{e}}\left( \dfrac{a}{b} \right)$ therefore, on using the property, we get:
\[\Rightarrow \left[ {{\log }_{e}}\left( \dfrac{x+\alpha }{x+\alpha +1} \right) \right]_{\alpha }^{\alpha +1}={{\log }_{e}}\left( \dfrac{9}{8} \right)\]
Now on substituting the value of the limits, we get:
\[\Rightarrow \left[ {{\log }_{e}}\left( \dfrac{\left( \alpha +1 \right)+\alpha }{\left( \alpha +1 \right)+\alpha +1} \right) \right]-\left[ {{\log }_{e}}\left( \dfrac{\alpha +\alpha }{\alpha +\alpha +1} \right) \right]={{\log }_{e}}\left( \dfrac{9}{8} \right)\]
On opening the brackets, we get:
\[\Rightarrow \left[ {{\log }_{e}}\left( \dfrac{\alpha +1+\alpha }{\alpha +1+\alpha +1} \right) \right]-\left[ {{\log }_{e}}\left( \dfrac{\alpha +\alpha }{\alpha +\alpha +1} \right) \right]={{\log }_{e}}\left( \dfrac{9}{8} \right)\]
On adding the similar terms, we get:
\[\Rightarrow \left[ {{\log }_{e}}\left( \dfrac{2\alpha +1}{2\alpha +2} \right) \right]-\left[ {{\log }_{e}}\left( \dfrac{2\alpha }{2\alpha +1} \right) \right]={{\log }_{e}}\left( \dfrac{9}{8} \right)\]
Now on applying the property of logarithm ${{\log }_{e}}a-{{\log }_{e}}b={{\log }_{e}}\left( \dfrac{a}{b} \right)$ on the above expression, we get:
\[\Rightarrow {{\log }_{e}}\left( \dfrac{\dfrac{2\alpha +1}{2\alpha +2}}{\dfrac{2\alpha }{2\alpha +1}} \right)={{\log }_{e}}\left( \dfrac{9}{8} \right)\]
On simplifying the fraction, we get:
\[\Rightarrow {{\log }_{e}}\left( \dfrac{2\alpha +1}{2\alpha +2}\times \dfrac{2\alpha +1}{2\alpha } \right)={{\log }_{e}}\left( \dfrac{9}{8} \right)\]
On simplifying terms by multiplying, we get:
\[\Rightarrow {{\log }_{e}}\left( \dfrac{{{\left( 2\alpha +1 \right)}^{2}}}{2\alpha \left( 2\alpha +2 \right)} \right)={{\log }_{e}}\left( \dfrac{9}{8} \right)\]
on using the expansion formula ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ on the numerator, we get:
\[\Rightarrow {{\log }_{e}}\left( \dfrac{4{{\alpha }^{2}}+4\alpha +1}{2\alpha \left( 2\alpha +2 \right)} \right)={{\log }_{e}}\left( \dfrac{9}{8} \right)\]
On multiplying the terms in the denominator, we get:
\[\Rightarrow {{\log }_{e}}\left( \dfrac{4{{\alpha }^{2}}+4\alpha +1}{4{{\alpha }^{2}}+4\alpha } \right)={{\log }_{e}}\left( \dfrac{9}{8} \right)\]
Now since logarithm is present on both the sides, we can remove it and write the terms as:
\[\Rightarrow \dfrac{4{{\alpha }^{2}}+4\alpha +1}{4{{\alpha }^{2}}+4\alpha }=\dfrac{9}{8}\]
On taking $4$ common from the denominator, we get:
\[\Rightarrow \dfrac{4{{\alpha }^{2}}+4\alpha +1}{4\left( {{\alpha }^{2}}+\alpha \right)}=\dfrac{9}{8}\]
On simplifying, we get:
\[\Rightarrow \dfrac{4{{\alpha }^{2}}+4\alpha +1}{\left( {{\alpha }^{2}}+\alpha \right)}=\dfrac{9}{2}\]
On cross multiplying, we get:
\[\Rightarrow 2\times \left( 4{{\alpha }^{2}}+4\alpha +1 \right)=9\times \left( {{\alpha }^{2}}+\alpha \right)\]
On multiplying the terms, we get:
\[\Rightarrow 8{{\alpha }^{2}}+8\alpha +2=9{{\alpha }^{2}}+9\alpha \]
On transferring the like terms across the $=$ , we get:
\[\Rightarrow 9{{\alpha }^{2}}-8{{\alpha }^{2}}+9\alpha -8\alpha -2=0\]
On simplifying, we get:
\[\Rightarrow {{\alpha }^{2}}+\alpha -2=0\]
Now the expression can be written in the factorized format as:
\[\Rightarrow \left( \alpha +2 \right)\left( \alpha -1 \right)=0\]
This implies that $\alpha =-2$ or $\alpha =1$.
Now from the list of given options, we have the option for $\alpha =-2$ and not $\alpha =-1$ therefore, the correct option is $\left( D \right)$, which is the required solution.

Note: It is to be remembered that in this question we are solving a definite integral, which has a limit value. There also exists indefinite integrals which do not have any limit to them. The log which we have used in the question is log to the base $e$. This is also written as $\ln $. The other commonly used base is log to the base $10$.