SATHEE: Indefinite Integration 1 Question 18

Indefinite Integration 1 Question 18

19. The integral $\int_{2}^{4} \frac{\log x^{2}}{\log x^{2} + \log (36 - 12 x + x^{2})} d x$ is equal to

(2015, Main)

(a) 2

(b) 4

(d) 6

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Solution:

PLAN Apply the property

$\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x and then add.$

Let $I = \int_{2}^{4} \frac{\log x^{2}}{\log x^{2} + \log (36 - 12 x + x^{2})} d x$

$\begin{aligned} = \int_{2}^{4} \frac{2 \log x}{2 \log x + \log (6 - x)^{2}} d x \\ = \int_{2}^{4} \frac{2 \log x d x}{2 [\log x + \log (6 - x)]} \end{aligned}$

$\begin{aligned} \Rightarrow I & = \int_{2}^{4} \frac{\log x d x}{[\log x + \log (6 - x)]} \\ \Rightarrow I & = \int_{2}^{4} \frac{\log (6 - x)}{\log (6 - x) + \log x} d x \\ ∵ \int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x \end{aligned}$

On adding Eqs. (i) and (ii), we get

$\begin{aligned} 2 I = \int_{2}^{4} \frac{\log x + \log (6 - x)}{\log x + \log (6 - x)} d x \\ \Rightarrow 2 I & = \int_{2}^{4} d x = [x]_{2}^{4} \Rightarrow 2 I = 2 \\ \Rightarrow 2 I & = 2 \Rightarrow I = 1 \end{aligned}$

Indefinite Integration 1 Question 18

19. The integral $\int_{2}^{4} \frac{\log x^{2}}{\log x^{2} + \log (36 - 12 x + x^{2})} d x$ is equal to

Solution:

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Indefinite Integration 1 Question 18

19. The integral ∫24log⁡x2log⁡x2+log⁡(36−12x+x2)dx is equal to

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

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NCERT Exercise Video Solution

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19. The integral $\int_{2}^{4} \frac{\log x^{2}}{\log x^{2} + \log (36 - 12 x + x^{2})} d x$ is equal to