Question

If \(\int \sqrt{e^{x}-1}dx=a\sqrt{e^{x}-1}+b\tan ^{-1}(\sqrt{e^{x}-1})+c\), then

• A. \(a+b=0\)
• B. \(a-b=0\)
• C. \(a=2b\)
• D. \(2a=b\)

Hint:

Substitute t equal to the square root of e to the x minus 1 and compare coefficients.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
Put \(t=\sqrt{e^{x}-1}\). Then \(e^{x}=1+t^{2}\) and \(dx=\frac{2t}{1+t^{2}}dt\).

Step 2: Integrate
\(\int t\cdot\frac{2t}{1+t^{2}}dt=2\int\left(1-\frac{1}{1+t^{2}}\right)dt=2t-2\tan^{-1}t+c\).

Step 3: Compare
Matching with \(a\sqrt{e^{x}-1}+b\tan^{-1}\sqrt{e^{x}-1}\) gives \(a=2\) and \(b=-2\), so \(a+b=0\).

Final Answer:
The relation is \(a+b=0\). \[ \boxed{\text{(A) }a+b=0} \]