Question

The integral ${\int }_2^4\frac{\log x^2}{\log x^2+\log (36-12x+x^2)}dx$ is equal to :

• A. $2$
• B. $4$
• C. $1$
• D. $6$

Answer 1

Answer: (C)

$1$

We have,

${\int }_2^4\frac{\log x^2}{\log x^2+\log (36-12x+x^2)}dx$

So,

We know that,

${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$

$I={\int }_2^4\frac{\log x^2}{\log x^2+\log {(6-x)}^2}dx$

$\Rightarrow I={\int }_2^4\frac{2\log x}{2\log x+2\log (6-x)}dx$

$\Rightarrow I={\int }_2^4\frac{\log x}{\log x+\log (6-x)}dx......\left(1\right)$

Using property and we get,

$\Rightarrow I={\int }_2^4\frac{\log (6-x)}{\log x+\log (6-x)}dx......\left(2\right)$

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

$2I={\int }_2^4\frac{\log x+\log (6-x)}{\log x+\log (6-x)}dx$

$2I={\int }_2^{4}1dx$

$2I={{\left[x\right]}_2}^4$

$2I=4-2$

$2I=2$

$I=1$

Hence, this is the answer.