Question

Find the integration of the given function:
$\begin{aligned} & \int{\dfrac{dx}{x\left( x-2 \right)\left( x-4 \right)}}= \\\ & \\\ \end{aligned}$
$\left( A \right)\dfrac{1}{8}\log \left| x \right|-\dfrac{1}{4}\log \left| x-2 \right|-\dfrac{1}{8}\log \left| x-4 \right|+c$
$\left( B \right)\dfrac{1}{8}\log \left| x \right|-\dfrac{1}{4}\log \left| x-2 \right|+\dfrac{1}{8}\log \left| x-4 \right|+c$
$\left( C \right)\dfrac{1}{8}\log \left| x \right|+\dfrac{1}{4}\log \left| x-2 \right|-\dfrac{1}{8}\log \left| x-4 \right|+c$
$\left( D \right)\dfrac{1}{8}\log \left| x \right|+\dfrac{1}{4}\log \left| x-2 \right|+\dfrac{1}{8}\log \left| x-4 \right|+c$

Answer 1

In such a question, first we have to check if the polynomial integral is a proper or improper fraction. If proper then decompose it into suitable partial fractions and integrate it..

Complete step-by-step answer:
Here degree of denominator is greater than degree of numerator, such that the given integrand is a proper fraction or rational function.
Now we have to decompose into suitable partial fraction
$\dfrac{1}{x\left( x-2 \right)\left( x-4 \right)}=\dfrac{A}{x}+\dfrac{B}{x-2}+\dfrac{C}{x-4}$ …………………………….(i)
$A,B\,and\,C$ are constants.
Now, to calculate the left part as same as fraction, we get
$\dfrac{1}{x\left( x-2 \right)\left( x-4 \right)}=\dfrac{A\left( x-2 \right)\left( x-4 \right)+B\left( x \right)\left( x-4 \right)+C\left( x \right)\left( x-2 \right)}{x\left( x-2 \right)\left( x-4 \right)}$
The denominator of both sides are equal, then it will be canceled
$1=A\left( x-2 \right)\left( x-4 \right)+B\left( x \right)\left( x-4 \right)+C\left( x \right)\left( x-2 \right)$…………………………………(ii)
Now we have to find the value of $A,B\,and\,C$.
To find the value of constants , we have to suppose some values for $x$ and put these values in equation (ii), to get the required values of$A,B\,and\,C$.
On putting $x=0$ in equation (ii), we get
$1=A\left( 0-2 \right)\left( 0-4 \right)+B\left( 0 \right)\left( 0-4 \right)+C\left( 0 \right)\left( 0-2 \right)$
$0\,$multiply with any number the whole term becomes $0\,$.
$1=A\left( -2 \right)\left( -4 \right)$
$1=A\times 8$
$A =\dfrac{1}{8}$
Now for the value of $B\,$, we have to put $x=2$
$1=A\left( 2-2 \right)\left( 2-4 \right)+B\left( 2 \right)\left( 2-4 \right)+C\left( 2 \right)\left( 2-2 \right)$
$1=A\times 0\left( 2-4 \right)+B\left( 2 \right)\left( -2 \right)+C\left( 2 \right)\times 0$
$1=B\left( -4 \right)$
$B=-\dfrac{1}{4}$
To find the value $\,C$, we have to put $x=4$
$1=A\left( 4-2 \right)\left( 4-4 \right)+B\left( 4 \right)\left( 4-4 \right)+C\left( 4 \right)\left( 4-2 \right)$
$1=A\left( 4-2 \right)\times 0+B\left( 4 \right)\times 0+C\left( 4 \right)\times 2$
$1=C\times 8$
$C=\dfrac{1}{8}$
On putting all these values of $A,B\,and\,C$ in equation (i), we get
$\dfrac{1}{x\left( x-2 \right)\left( x-4 \right)}=\dfrac{1}{8x}-\dfrac{1}{4(x-2)}+\dfrac{1}{8(x-4)}$
On integrating both sides,
$\int{(}\dfrac{1}{x\left( x-2 \right)\left( x-4 \right)})dx=\int{(}\dfrac{1}{8x}-\dfrac{1}{4(x-2)}+\dfrac{1}{8(x-4)})dx$
$\int{(}\dfrac{1}{x\left( x-2 \right)\left( x-4 \right)})dx=\int{{}}\dfrac{1}{8x}dx-\int{{}}\dfrac{1}{4(x-2)}dx+\int{{}}\dfrac{1}{8(x-4)}dx$………..(iii)
We know that from the formula of integration,
$\int{\dfrac{1}{x}}dx=\log \left| x \right|+c$
On applying this formula ,we get
$\int{(}\dfrac{1}{x\left( x-2 \right)\left( x-4 \right)})dx=\dfrac{1}{8}\log \left| x \right|-\dfrac{1}{4}\log \left| x-2 \right|+\dfrac{1}{8}\log \left| x-4 \right|+c$
Where c is a constant of integration.

So, the correct answer is “Option B”.

Additional Information: In such types of problems, if the degree of numerator is greater than the degree of denominator, we have to convert it into proper fraction by dividing the numerator by denominator ,otherwise integration is not possible.

Note: We can solve this question by other method like we can convert it into a polynomial in $x$ and equate the coefficients of ${{x}^{3}},{{x}^{2}},x$ and constant terms from both sides to get equations in $A,B,C.$.On solving these equation , we get values of $A,B,C.$