Question

The value of the definite integral \(\int _0^{log_e5}\frac{e^x\sqrt{e^x-1}}{e^x+3}\,dx\) is equal to...

• A. \(5+π\)
• B. \(5-π\)
• C. \(4+π\)
• D. \(4-π\)

Hint:

Substitute t = sqrt(e^x - 1) to remove the square root.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Put \(t = \sqrt{e^x - 1}\), so \(t^2 = e^x - 1\) and \(2t\,dt = e^x\,dx\). Also \(e^x + 3 = t^2 + 4\).

Step 2: Key Formula or Approach:
Limits: \(x = 0 \Rightarrow t = 0\); \(x = \log_e 5 \Rightarrow t = \sqrt{5 - 1} = 2\).

Step 3: Detailed Explanation:
\[ I = \int_0^2\frac{t\cdot 2t\,dt}{t^2 + 4} = 2\int_0^2\frac{t^2}{t^2 + 4}dt = 2\int_0^2\left(1 - \frac{4}{t^2 + 4}\right)dt \]
\[ = 2\left[t - 2\tan^{-1}\frac t2\right]_0^2 = 2\left[2 - 2\cdot\frac\pi4\right] = 4 - \pi \]

Final Answer:
The value is \(4 - \pi\), option (D). \[ \boxed{4-\pi} \]