Question

\(\int _0^{log5}\frac{e^x\sqrt{e^x-1}}{e^x+3}\,dx =\)

• A. \(2-\frac{π}{4}\)
• B. \(4-\frac{π}{4}\)
• C. \(2-π\)
• D. \(4-π\)

Hint:

Let u = sqrt(e^x - 1) so that e^x dx = 2u du.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Remove the square root with \(u = \sqrt{e^x - 1}\), so \(e^x = u^2 + 1\) and \(e^xdx = 2u\,du\).

Step 2: Change the limits:
At \(x = 0\): \(u = 0\). At \(x = \log5\): \(u = \sqrt{4} = 2\).
Also \(e^x + 3 = u^2 + 4\).

Step 3: Rewrite and integrate:
\[ I = \int_0^2\frac{u\cdot2u\,du}{u^2 + 4} = 2\int_0^2\frac{u^2}{u^2+4}du = 2\int_0^2\left(1 - \frac{4}{u^2+4}\right)du \]
\[ = 2\left[u - 2\tan^{-1}\frac u2\right]_0^2 = 2\left[2 - 2\cdot\frac\pi4\right] \]

Step 4: Result:
\[ I = 4 - \pi \]
Options (A) and (B) have \(\pi/4\) instead of \(\pi\), and (C) is \(2 - \pi\), which is negative and cannot be right for a positive integrand.

Final Answer:
The substitution gives 4 - pi. \[ \boxed{\text{(D) }4-\pi} \]