Find \(\int_1^2 \frac{12 \,logx}{x} \,dx\).

(a) -12 log2

(b) 24 log2

(c) 12 log2

(d) 24 log4

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This key question is from Evaluation of Definite Integrals by Substitution in portion Integrals of Mathematics – Class 12

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Right choice is (b) 24 log2

To explain I would say: I=\int_1^2 \frac{12 logx}{x} \,dx

Let logx=t

Differentiating w.r.t x, we get

\frac{1}{x} \,dx=dt

The new limits

When x=1,t=0

When x=2,t=log2

\int_1^2 \frac{12 logx}{x} dx=12\int_0^{log2} \,t \,dt

=12[t^2]_0^{log2}=12((log2)^2-0)

=12 log4=24 log2(∵(log2)^2=log2.log2=log4=2 log2)