Question

Integration $\int{\dfrac{xdx}{\left( x-1 \right)\left( x-2 \right)}}$ equals
A. $\log \left| \dfrac{{{\left( x-1 \right)}^{2}}}{x-2} \right|+C$
B. $\log \left| \dfrac{{{\left( x-2 \right)}^{2}}}{x-1} \right|+C$
C. $\log \left| \dfrac{{{\left( x-2 \right)}^{2}}}{x-2} \right|+C$
D. $\log \left| \left( x-1 \right)\left( x-2 \right) \right|+C$

Answer 1

Here we have been given an integral which we have to solve. Firstly as we can see that the power in the denominator is bigger than the power in the numerator so we will use a partial fraction to simplify the term inside the integral. Then we will separately find the integration of the two terms obtained by using the basic integration formula. Finally we will simplify the answer obtained and get our desired answer.

Complete step-by-step solution:
We have to solve the below integral,
$\int{\dfrac{xdx}{\left( x-1 \right)\left( x-2 \right)}}$….$\left( 1 \right)$
We will use partial fraction where we will simplify the term inside the integral sign as follows:
So let,
$\dfrac{x}{\left( x-1 \right)\left( x-2 \right)}=\dfrac{A}{x-1}+\dfrac{B}{x-2}$….$\left( 2 \right)$
On taking the L.C.M on right hand side we get,
$\Rightarrow \dfrac{x}{\left( x-1 \right)\left( x-2 \right)}=\dfrac{A\left( x-2 \right)+B\left( x-1 \right)}{\left( x-1 \right)\left( x-2 \right)}$
Cancel the denominator on both sides as they are same and simplifying the numerator we get,
$\Rightarrow x=Ax-2A+Bx-B$
$\Rightarrow x=x\left( A+B \right)+\left( -2A-B \right)$….$\left( 3 \right)$
Now compare the coefficient of $x$ both sides we get,
$A+B=1$
$A=1-B$….$\left( 4 \right)$
Compare the coefficient of the constant on both side of equation (3) we get,
$-2A-B=0$
Substitute the value from equation (4) above we get,
$\Rightarrow -2\left( 1-B \right)-B=0$
$\Rightarrow -2+2B-B=0$
Simplifying further,
$\Rightarrow -2+B=0$
$\Rightarrow B=2$
Replace the value in equation (4) we get,
$A=1-2$
$\Rightarrow A=-1$
So we get $A=-1$ and $B=2$ substitute the value in equation (2) we get,
$\dfrac{x}{\left( x-1 \right)\left( x-2 \right)}=\dfrac{-1}{x-1}+\dfrac{2}{x-2}$
Put the value in equation (1),
$\int{\dfrac{-1}{x-1}+\dfrac{2}{x-2}\,dx}$….$\left( 5 \right)$
We know,
$\int{\dfrac{1}{x}dx=\log x+C}$
Using the formula in equation (5) we get,
$\int{\dfrac{-1}{x-1}+\dfrac{2}{x-2}\,dx}=-1\log \left( x-1 \right)+2\log \left( x-2 \right)+C$
Now as we know logarithm property that $a\log b=\log {{b}^{a}}$ and $\log a-\log b=\log \dfrac{a}{b}$ using them above,
$\Rightarrow \int{\dfrac{-1}{x-1}+\dfrac{2}{x-2}\,dx}=-\log \left( x-1 \right)+\log {{\left( x-2 \right)}^{2}}+C$
$\Rightarrow \int{\dfrac{-1}{x-1}+\dfrac{2}{x-2}\,dx}=\log \left( \dfrac{{{\left( x-2 \right)}^{2}}}{\left( x-1 \right)} \right)+C$
Hence the correct option is (B).

Note: Integration is among two major topics of calculus in mathematics. It denotes the summation of discrete data. It is used to find the function which will describe the area, volume that occur due to collection of small data which we can’t measure in separately. Integration is used to find the problem function when the derivative is given and also to find the area bounded by the graph of a function under certain constraints.