Question

$\int_2^4\dfrac{logx^{2}}{logx^{2}+log(x^{2}-12x+36)} dx$

Answer 1

$\large\underline{\sf{Solution-}}$

Given integral is

$\rm \: \displaystyle\int_{2}^{4}\rm \frac{log {x}^{2} }{log {x}^{2} + log( {x}^{2} - 12x + 36)} \: dx$

Let assume that

$\rm \: I \: = \: \displaystyle\int_{2}^{4}\rm \frac{log {x}^{2} }{log {x}^{2} + log( {x}^{2} - 12x + 36)} \: dx$

can be rewritten as

$\rm \: I \: = \: \displaystyle\int_{2}^{4}\rm \frac{log {x}^{2} }{log {x}^{2} + log {(x - 6)}^{2} } \: dx - - - (1)$

We know,

$\boxed{ \rm{ \:\displaystyle\int_{a}^{b}\rm f(x)dx \: = \: \displaystyle\int_{a}^{b}\rm f(a + b - x)dx \: }}$

So, using this property, we get

$\rm \: I \: = \: \displaystyle\int_{2}^{4}\rm \frac{log {(2 + 4 - x)}^{2} }{log {(2 + 4 - x)}^{2} + log {(6 - 2 - 4 + x)}^{2} } \: dx$

$\rm \: I \: = \: \displaystyle\int_{2}^{4}\rm \frac{log {(6- x)}^{2} }{log {(6- x)}^{2} + log {x}^{2} } \: dx - - - (2)$

On adding equation (1) and (2), we get

$\rm \: 2I \: = \: \displaystyle\int_{2}^{4}\rm \frac{log {x}^{2} + log {(6- x)}^{2} }{log {(6- x)}^{2} + log {x}^{2} } \: dx$

$\rm \: 2I \: = \: \displaystyle\int_{2}^{4}\rm 1 \: dx$

$\rm \: 2I \: = \: \bigg(x\bigg) _{2}^{4}$

$\rm \: 2I \: = \: 4 - 2$

$\rm \: 2I \: = \: 2$

$\rm\implies \:I = 1$

Hence,

$\rm\implies \boxed{\rm{\displaystyle\int_{2}^{4}\rm \frac{log {x}^{2} }{log {x}^{2} + log( {x}^{2} - 12x + 36)}dx = 1 \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

$\boxed{ \rm{ \:\displaystyle\int_{a}^{b}\rm f(x)dx \: = \: \displaystyle\int_{a}^{b}\rm f(y)dy \: }}$

$\boxed{ \rm{ \:\displaystyle\int_{a}^{b}\rm f(x)dx \: = \: - \: \displaystyle\int_{b}^{a}\rm f(x)dx \: }}$

$\boxed{ \rm{ \:\displaystyle\int_{0}^{a}\rm f(x)dx \: = \: \displaystyle\int_{0}^{a}\rm f(a - x)dx \: }}$

$\boxed{ \rm{ \:\displaystyle\int_{ - a}^{a}\rm f(x)dx \: = \:2 \displaystyle\int_{0}^{a}\rm f(x)dx \: \: if \: f( - x) = f(x)}}$

$\boxed{ \rm{ \:\displaystyle\int_{ - a}^{a}\rm f(x)dx \: = \:0 \: \: if \: f( - x) = - \: f(x)}}$

$\boxed{ \rm{ \:\displaystyle\int_{0}^{2a}\rm f(x)dx \: = \:0 \: \: if \: f(2a - x) = - \: f(x)}}$

$\boxed{ \rm{ \:\displaystyle\int_{0}^{2a}\rm f(x)dx \: = \:2\displaystyle\int_{0}^{a}\rm f(x)dx \: \: if \: f(2a - x) =\: f(x)}}$

Answer 2

Question :-

$\\ \rm\int_2^4\dfrac{logx^{2}}{logx^{2}+log(x^{2}-12x+36)} \: \: dx$

$\rule{300pt}{0.1pt}$

$\color{violet} \underline{ \begin{array}{ || |l| || } \hline \color{magenta} \\ \hline \boxed{ \text{ \tt \: Solution:-} } \end{array}}$

$\[ \begin{array}{l} \displaystyle\rm I=\int_{2}^{4} \frac{\log x^{2}}{\log x^{2}+\log \left(36-12 x+x^{2}\right)} d x \\\\ \\ \displaystyle\rm I=\frac{2}{2} \int_{2}^{4} \frac{\log x}{\log x+\log (6-x)} d x \\\\ \\ \displaystyle\rm I=\int_{2}^{4} \frac{\log (6-x)}{\log (6-x)+\log x} d x \qquad \qquad\left\{\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\right\} \end{array} \]$

Adding (i) and (ii)

$\begin{array}{l} \displaystyle\ \rm\[ 2 I=\int_{2}^{4} \frac{\log x+\log (6-x)}{\log x+\log (6-x)} d x \\ \\ \displaystyle \rm 2 I=\int_{2}^{4}1 d x \\ \\ \pmb{\rm 2 I=2 }\] \end{array}$

Hence, I = 1