Question

If $\int_2^e \left[ \frac{1}{\log x} - \frac{1}{(\log x)^2} \right] dx = a + \frac{b}{\log 2}$, then

• A. $a = -e, b = 2$
• B. $a = e, b = -2$
• C. $a = e, b = 2$
• D. $a = -e, b = -2$

Hint:

Whenever you see an integral involving $1/(\log x)$ and its derivative, substituting $x = e^t$ instantly converts it into the famous cancellation format $\int e^t [f(t) + f'(t)] dt$.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to evaluate a definite integral involving logarithmic functions and find the corresponding values of constants $a$ and $b$.

Step 2: Key Formula or Approach:
Use the substitution method by setting $\log x = t$, which transforms the integral into the standard form $\int e^t [f(t) + f'(t)] dt = e^t f(t) + c$.

Step 3: Detailed Explanation:
Let $I = \int_2^e \left[ \frac{1}{\log x} - \frac{1}{(\log x)^2} \right] dx$.
Substitute $\log x = t$. This gives $x = e^t$, and differentiating gives $dx = e^t dt$.
Now change the integration limits:
When $x = 2$, $t = \log 2$.
When $x = e$, $t = \log e = 1$.
Substitute these into the integral:
$I = \int_{\log 2}^1 \left( \frac{1}{t} - \frac{1}{t^2} \right) e^t dt$
This perfectly matches the form $\int e^t [f(t) + f'(t)] dt$, where $f(t) = \frac{1}{t}$ and $f'(t) = -\frac{1}{t^2}$.
Therefore, the antiderivative is $e^t f(t) = e^t \cdot \frac{1}{t}$.
Applying the limits:
$I = \left[ \frac{e^t}{t} \right]_{\log 2}^1 = \left( \frac{e^1}{1} \right) - \left( \frac{e^{\log 2}}{\log 2} \right)$
Since $e^{\log 2} = 2$, we get:
$I = e - \frac{2}{\log 2}$
Comparing this result with the given expression $a + \frac{b}{\log 2}$, we find $a = e$ and $b = -2$.

Step 4: Final Answer:
The values are $a = e, b = -2$, matching option (B).