Question

The integral $\int\limits_{2}^{4}{\dfrac{\log {{x}^{2}}}{\log {{x}^{2}}+\log \left( 36-12x+{{x}^{2}} \right)}dx}$ is equal to
(a) 3
(b) 2
(c) 1
(d) 0

Answer 1

Start by letting the integral to be I and go for simplification of the definite integral using the property $\int\limits_{a}^{b}{f\left( x \right)dx}=\int\limits_{a}^{b}{f\left( a+b-x \right)dx}$ . After simplification, add both the forms of integration and solve to get the answer.

Complete step-by-step answer:
Before starting the solution, let us discuss the important properties of definite integration.
Some important properties are:
$\int\limits_{a}^{b}{f\left( x \right)dx}=\int\limits_{a}^{b}{f\left( a+b-x \right)dx}$
$\int\limits_{a}^{b}{f\left( x \right)dx}=\int\limits_{a}^{c}{f\left( x \right)dx+}\int\limits_{c}^{b}{f\left( x \right)dx}$
Now let us start with the integral given in the above question.
$I=\int\limits_{2}^{4}{\dfrac{\log {{x}^{2}}}{\log {{x}^{2}}+\log \left( 36-12x+{{x}^{2}} \right)}dx}$
$\Rightarrow I=\int\limits_{2}^{4}{\dfrac{\log {{x}^{2}}}{\log {{x}^{2}}+\log \left( {{6}^{2}}-2\times 6\times x+{{x}^{2}} \right)}dx}$
Now, we know that ${{\left( 6-x \right)}^{2}}=36-12x+{{x}^{2}}$ according to the formula ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ .
$I=\int\limits_{2}^{4}{\dfrac{\log {{x}^{2}}}{\log {{x}^{2}}+\log {{\left( 6-x \right)}^{2}}}dx}............(i)$
Using the property $\int\limits_{a}^{b}{f\left( x \right)dx}=\int\limits_{a}^{b}{f\left( a+b-x \right)dx}$ , we get
$I=\int\limits_{2}^{4}{\dfrac{\log {{\left( 2+4-x \right)}^{2}}}{\log {{\left( 2+4-x \right)}^{2}}+\log {{\left( 6-\left( 2+4-x \right) \right)}^{2}}}dx}$
$\Rightarrow I=\int\limits_{2}^{4}{\dfrac{\log {{\left( 6-x \right)}^{2}}}{\log {{\left( 6-x \right)}^{2}}+\log {{x}^{2}}}dx}.........(ii)$
Now, we will add equation (i) and equation (ii). On doing so, we get
$2I=\int\limits_{2}^{4}{\dfrac{\log {{\left( 6-x \right)}^{2}}}{\log {{\left( 6-x \right)}^{2}}+\log {{x}^{2}}}dx}+\int\limits_{2}^{4}{\dfrac{\log {{x}^{2}}}{\log {{\left( 6-x \right)}^{2}}+\log {{x}^{2}}}dx}$
$\Rightarrow 2I=\int\limits_{2}^{4}{\dfrac{\log {{\left( 6-x \right)}^{2}}+\log {{x}^{2}}}{\log {{\left( 6-x \right)}^{2}}+\log {{x}^{2}}}dx}$
As denominator and numerator are equal, so they get cancelled. On doing so, we get
$2I=\int\limits_{2}^{4}{dx}$
$\Rightarrow 2I=\left. x \right]_{2}^{4}=4-2=2$
$\Rightarrow I=1$

So, the correct answer is “Option C”.

Note: Remember that in case of definite integral whenever you try to express a function in terms of some other variable, i.e., use substitution, the limits also change. However, for when you use properties of definite integral the limits don't change. The general mistake a student makes is they substitute x but forget to change the limits or use definite integration related properties and change the limits, leading to errors. Also, you need to remember all the basic formulas that we use for indefinite integrals as they are used in definite integrations as well.