Question

\(\int _0^{log10}[\frac{e^x\sqrt{e^x-1}}{e^x+8}]dx =\)

• A. \(\frac{3}{2}(4+π)\)
• B. \(\frac{3}{2}(4-π)\)
• C. \(\frac{3}{4}(4-π)\)
• D. \(\frac{3}{4}(4+π)\)

Hint:

Put t = root of (e^x - 1), so e^x dx = 2t dt and the limits become 0 to 3.

Answer 1

The Correct Option is B

Step 1: Choose a substitution:
The root \(\sqrt{e^x - 1}\) suggests \(t = \sqrt{e^x - 1}\). Then \(e^x = t^2 + 1\) and, by differentiating, \(e^x\,dx = 2t\,dt\).

Step 2: Change the limits:
At \(x = 0\): \(t = \sqrt{1-1} = 0\). At \(x = \log 10\): \(t = \sqrt{10-1} = 3\). So the new limits are \(0\) to \(3\).

Step 3: Rewrite the integral:
The numerator \(e^x\sqrt{e^x-1}\,dx\) becomes \(t\cdot 2t\,dt\), and the denominator \(e^x + 8 = t^2 + 9\). So
\[ I = \int_0^3 \frac{2t^2}{t^2+9}\,dt = 2\int_0^3\left(1 - \frac{9}{t^2+9}\right)dt \]

Step 4: Evaluate:

\[ I = 2[t]_0^3 - 18\cdot\frac{1}{3}\left[\tan^{-1}\frac{t}{3}\right]_0^3 = 6 - 6\left(\frac{\pi}{4} - 0\right) = 6 - \frac{3\pi}{2} \]
So \(I = \frac{3}{2}(4 - \pi)\).

Step 5: Why the other options are wrong:
Options with \(4 + \pi\) would come from adding the arctangent part instead of subtracting it. The minus sign comes from \(t^2 = (t^2+9) - 9\). The factor \(\frac{3}{4}\) in (C) and (D) would need \(6 - \frac{3\pi}{2}\) to be halved, but the factor \(2\) in \(2t\,dt\) was already used. A quick check: \(\frac{3}{2}(4-\pi) \approx 1.29\), which fits an integrand that stays below about \(1.7\) on a range of length about \(2.3\).

Final Answer:
The value of the integral is \(\frac{3}{2}(4-\pi)\), which is option (B). \[ \boxed{\frac{3}{2}(4-\pi)} \]