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: Try t = e^x:
Let $t = e^x$, $dt = e^xdx$. Limits $1$ to $5$. $I = \int_1^5\dfrac{\sqrt{t-1}}{t+3}dt$.

Step 2: Substitute again:
Put $t - 1 = v^2$, $dt = 2v\,dv$, limits $0$ to $2$: $I = \int_0^2\dfrac{2v^2}{v^2+4}dv$.

Step 3: Evaluate:
$2\int_0^2\left(1 - \dfrac4{v^2+4}\right)dv = 2\left[2 - 4\cdot\tfrac12\cdot\tfrac\pi4\right] = 4 - \pi$. This is option (D).

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