Question

Find the value of \[\int\limits_{\rm{5}}^{{\rm{10}}} {\dfrac{{\rm{1}}}{{\left[ {\left( {{\rm{x - 1}}} \right)\left( {{\rm{x - 2}}} \right)} \right]}}{\rm{dx}}} \] .
A. \[\log \left( {\dfrac{{27}}{{32}}} \right)\]
B. \[\log \left( {\dfrac{{32}}{{27}}} \right)\]
C. \[\log \left( {\dfrac{8}{9}} \right)\]
D. \[\log \left( {\dfrac{3}{4}} \right)\]

Answer 1

Hint: Convert the given integral into partial fraction and then find the value of given variables and then substitute the values into the partial integral and then solve the integral further.

Formula used:
\[\int\limits_a^b {\left[ {f(x) + f(y)} \right]} = \int\limits_a^b {f(x) + \int\limits_a^b {f(y)} } \]
Logarithm formula, \[\ln a - \ln b = \ln \dfrac{a}{b}\]

Complete step by step solution:
The given integral is, \[\int\limits_5^{10} {\dfrac{{dx}}{{\left[ {\left( {x - 1} \right)\left( {x - 2} \right)} \right]}}} \]
Now, Converting the expression into partial fractions, we have:
\[\dfrac{1}{{\left( {x - 1} \right)\left( {x - 2} \right)}} = \dfrac{A}{{x - 1}}{\rm{ }} + {\rm{ }}\dfrac{B}{{x - 2}}\]
Where A and B are arbitrary constants,
\[\dfrac{1}{{\left( {x - 1} \right)\left( {x - 2} \right)}} = \dfrac{{A\left( {x - 2} \right) + B\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {x - 2} \right)}}\]
\[\dfrac{1}{{\left( {x - 1} \right)\left( {x - 2} \right)}} = \dfrac{{Ax - 2A + Bx - B}}{{\left( {x - 1} \right)\left( {x - 2} \right)}}\]
\[1 = Ax - 2A + Bx - B\]
\[1 = Ax + Bx - 2A - B\]
\[1 = (A + B)X + ( - 2A - B)\]
\[1 = (A + B)X + ( - 2A - B)\]
\[0 \cdot x + 1 = (A + B)X + ( - 2A - B)\]
Comparing both sides and we get:
\[A{\rm{ }} + {\rm{ }}B{\rm{ }} = {\rm{ }}0\]………………… (1)
\[ - 2A{\rm{ }} - {\rm{ }}B{\rm{ }} = 1\]………………………… (2)
Solving (1) and (2), We get:
\[A = {\rm{ }} - 1,B{\rm{ }} = 1\]
Further, putting the value of A and B, we get:
\[I{\rm{ }} = {\rm{ }}\int\limits_5^{10} {\left( {\dfrac{{ - 1}}{{x - 1}}{\rm{ }} + {\rm{ }}\dfrac{1}{{x - 2}}} \right)} dx\]
We know that, \[\int\limits_a^b {\left[ {f(x) + f(y)} \right]} = \int\limits_a^b {f(x) + \int\limits_a^b {f(y)} } \]
Now,
\[ = {\rm{ }}[ - \ln (x - 1){\rm{ }} + {\rm{ }}\ln (x - 2)]_5^{10}\]
\[ = {\rm{ }}\left[ {\ln \dfrac{{\left( {x - 2} \right)}}{{\left( {x - 1} \right)}}} \right]_5^{10}\]

\[ = \left[ {\ln \left( {\dfrac{{10 - 2}}{{10 - 1}}} \right) - \ln \left( {\dfrac{{5 - 2}}{{5 - 1}}} \right)} \right]\]
\[ = \ln \dfrac{8}{9} - \ln \dfrac{3}{4}\]
\[ = \ln \dfrac{{32}}{{27}}\]
The correct answer is option B.

Note Sometimes students make mistakes while starting with the integral as they start integration direct without breaking it into parts. This leads to a complex expression.