Question

If \[ \int \log x\sqrt{\left(\frac{\log x}{x}\right)^2+\frac1{x^2}}\,dx = \frac{f(x)}{3}\sqrt{1+(\log x)^2}+C \] and \(f(1)=1\), then \(f(e)=\) ?

• A. \(\frac23\)
• B. \(2\)
• C. \(\frac13\)
• D. \(6\)

Hint:

When both \(\log x\) and \(\frac1x\) appear together, try the substitution \(t=\log x\) immediately.

Answer 1

The Correct Option is B

Concept: Whenever expressions involving \(\log x\) and \(\frac1x\) appear together, the substitution \[ t=\log x \] usually simplifies the integral. The identity \[ \sqrt{\left(\frac{\log x}{x}\right)^2+\frac1{x^2}} = \frac{\sqrt{1+(\log x)^2}}{x} \] is the key observation.

Step 1: Simplify the integrand.
Given \[ I= \int \log x \sqrt{\left(\frac{\log x}{x}\right)^2+\frac1{x^2}} \,dx. \] Taking \(\frac1{x^2}\) common inside the square root, \[ I= \int \frac{\log x}{x} \sqrt{1+(\log x)^2} \,dx. \]

Step 2: Use substitution \(t=\log x\).
Let \[ t=\log x. \] Then \[ dt=\frac{dx}{x}. \] Therefore, \[ I = \int t\sqrt{1+t^2}\,dt. \]

Step 3: Integrate.
Let \[ u=1+t^2. \] Then \[ du=2t\,dt. \] Hence \[ I = \frac12\int u^{1/2}du. \] \[ = \frac12\cdot\frac23u^{3/2}+C. \] \[ = \frac13(1+t^2)^{3/2}+C. \] Substituting back, \[ I = \frac13\Big(1+(\log x)^2\Big)^{3/2}+C. \]

Step 4: Compare with the given form.
Given \[ I = \frac{f(x)}3 \sqrt{1+(\log x)^2}+C. \] Thus \[ f(x) = 1+(\log x)^2. \]

Step 5: Find \(f(e)\).
Since \[ \log e=1, \] \[ f(e) = 1+1^2 = 2. \] Therefore, \[ \boxed{2}. \]