Question

If \(\int _2^e[\frac{1}{logx}-\frac{1}{(logx)^2}]dx = a+\frac{b}{log2}\), then values of \(a\) and \(b\) are

• A. \(a = e,b = 2\)
• B. \(a = -e,b = 2\)
• C. \(a = e,b = -2\)
• D. \(a = -e,b = -2\)

Hint:

The integrand is the derivative of x / ln x.

Answer 1

The Correct Option is C

Step 1: Spot the antiderivative:
Consider \(\dfrac{d}{dx}\left(\dfrac{x}{\ln x}\right)=\dfrac{\ln x\cdot1-x\cdot\frac1x}{(\ln x)^2}=\dfrac{\ln x-1}{(\ln x)^2}=\dfrac{1}{\ln x}-\dfrac{1}{(\ln x)^2}\). So the integrand is exactly this derivative.

Step 2: Apply the limits:
\[ \left[\dfrac{x}{\ln x}\right]_2^e=\dfrac{e}{\ln e}-\dfrac{2}{\ln2}=e-\dfrac{2}{\ln2} \]

Step 3: Compare with a + b/log2:
So \(a=e\) and \(b=-2\). Option C.

Step 4: Why the other options are wrong.
Options A and B give \(b=+2\), and options B and D give \(a=-e\). The upper limit gives a positive \(e\) and the lower limit gives \(-\dfrac2{\ln2}\).

Final Answer:
a = e and b = -2. \[ \boxed{\text{(C) }a=e,\ b=-2} \]