Let ${I_1} = \mathop \smallint \limits_1^2 \,\,\frac{1}{{\sqrt {1\,\, + \,\,{x^2}} }}\,\,dx$ and ${I_2} = \mathop \smallint \limits_1^2 \,\,\frac{1}{x}\,\,dx$. Then

JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

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Let

{I_1} = \int \limits_1^2 \,\,\frac{1}{{\sqrt {1\,\, + \,\,{x^2}} }}\,\,dx

and

{I_2} = \int \limits_1^2 \,\,\frac{1}{x}\,\,dx

. Then

View answer and explanation

Correct option: B

{I_2} > {I_1}

Option B (I2 > I1) is correct because for x in [1, 2], 1/√(1+x²) < 1/x, implying that the integral I2 is greater than I1.

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