If ₂^{e}≤ft[1/ x-{1}{( x)²}] d x=a+b/ 2, then values of a and b are

Correct Answer is : a=e,b=−2a=e, b=-2a=e,b=−2

Correct Answer is : a=e,b=−2a=e, b=-2a=e,b=−2

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MHTCET 11 Apr 2026 Shift 2 Paper

Question : 131 of 150

Marks: +1, -0

\int_{2}^{e}\left[\frac{1}{\log x}-\frac{1}{(\log x)^{2}}\right] d x=a+\frac{b}{\log 2}

a

b

are

Solution:

Concept:

The integrand is the derivative of

\frac{x}{\log x}

, which allows direct integration using the fundamental theorem of calculus.

Explanation:

Start with

\frac{d}{dx}\left(\frac{x}{\log x}\right)

Using the quotient rule,

\frac{d}{dx}\left(\frac{x}{\log x}\right)=\frac{\log x \cdot 1 - x\cdot \frac{1}{x}}{(\log x)^2}=\frac{\log x - 1}{(\log x)^2}.

Rewrite this as,

\frac{\log x}{(\log x)^2}-\frac{1}{(\log x)^2}=\frac{1}{\log x}-\frac{1}{(\log x)^2}.

Therefore,

\int_{2}^{e}\left[\frac{1}{\log x}-\frac{1}{(\log x)^2}\right]dx = \left[\frac{x}{\log x}\right]_{2}^{e}.

Substitute the limits,

= \frac{e}{\log e}-\frac{2}{\log 2}=e-\frac{2}{\log 2},

since

\log e = 1

Comparing with

a+\frac{b}{\log 2}

, we get

a=e

and

b=-2

Answer:

a=e

b=-2

Option C.

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