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Grade:10

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
9 years ago
Ans:
Hello Student,
Please find answer to your question below

I = \int_{2}^{4}[log_{x}2- \frac{(log_{x}2)^{2}}{log2}]dx
I = \int_{2}^{4}[\frac{log2}{logx}- \frac{(log2)^{2}}{log2.(logx)^{2}}]dx
I = \int_{2}^{4}[\frac{log2.(logx)}{(logx)^2}- \frac{(log2)}{(logx)^{2}}]dx
I = log2\int_{2}^{4}[\frac{(logx)}{(logx)^2}- \frac{1}{(logx)^{2}}]dx
I = log2\int_{2}^{4}[\frac{(logx)-1}{(logx)^2}]dx
I = log2[\frac{x}{(logx)}]_{2}^{4}
I = log2[\frac{4}{(log4)}-\frac{2}{log2}]
I = log2[\frac{4}{2log2}-\frac{2}{log2}]
I = log2[\frac{2}{log2}-\frac{2}{log2}]
I = 0
Option (1) is correct.

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