Indefinite Integration 1 Question 18
19. The integral $\int _2^{4} \frac{\log x^{2}}{\log x^{2}+\log \left(36-12 x+x^{2}\right)} d x$ is equal to
(2015, Main)
(a) 2
(b) 4
(c) 1
(d) 6
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Solution:
- PLAN Apply the property
$$ \int _a^{b} f(x) d x=\int _a^{b} f(a+b-x) d x \text { and then add. } $$
Let $\quad I=\int _2^{4} \frac{\log x^{2}}{\log x^{2}+\log \left(36-12 x+x^{2}\right)} d x$
$$ \begin{aligned} & =\int _2^{4} \frac{2 \log x}{2 \log x+\log (6-x)^{2}} d x \\ & =\int _2^{4} \frac{2 \log x d x}{2[\log x+\log (6-x)]} \end{aligned} $$
$$ \begin{aligned} \Rightarrow \quad I & =\int _2^{4} \frac{\log x d x}{[\log x+\log (6-x)]} \\ \Rightarrow \quad I & =\int _2^{4} \frac{\log (6-x)}{\log (6-x)+\log x} d x \\ & \because \int _a^{b} f(x) d x=\int _a^{b} f(a+b-x) d x \end{aligned} $$
On adding Eqs. (i) and (ii), we get
$$ \begin{aligned} & 2 I=\int _2^{4} \frac{\log x+\log (6-x)}{\log x+\log (6-x)} d x \\ \Rightarrow \quad 2 I & =\int _2^{4} d x=[x] _2^{4} \Rightarrow 2 I=2 \\ \Rightarrow \quad 2 I & =2 \Rightarrow I=1 \end{aligned} $$
