Question

If \(f(x)=\frac{1}{\sqrt{x}},g(x)=3-2x-x^{2}\) then \(\int [(f\circ g)(e^{x})]\cdot e^{x}dx=\cdots \cdots \cdots\)

• A. \(\frac{1}{2}\sin ^{-1}\left(\frac{e^{x}+1}{2}\right)+c\)
• B. \(\sin ^{-1}\left(\frac{e^{x}-1}{4}\right)+c\)
• C. \(\sin ^{-1}\left(\frac{e^{x}+2}{2}\right)+c\)
• D. \(\sin ^{-1}\left(\frac{e^{x}+1}{2}\right)+c\)

Hint:

Put t = e^x and complete the square under the root to get an inverse sine form.

Answer 1

The Correct Option is D

Step 1: Differentiate Option (D)
Take $y=\sin^{-1}\dfrac{e^{x}+1}{2}$. Then $\dfrac{dy}{dx}=\dfrac{e^{x}/2}{\sqrt{1-\frac{(e^{x}+1)^{2}}{4}}}$.

Step 2: Simplify
$\dfrac{dy}{dx}=\dfrac{e^{x}}{\sqrt{4-(e^{x}+1)^{2}}}=\dfrac{e^{x}}{\sqrt{3-2e^{x}-e^{2x}}}$.

Step 3: Compare
This equals the integrand $(f\circ g)(e^{x})\cdot e^{x}$, so option (D) is correct.

Final Answer:
Differentiation of option (D) returns the integrand. \[ \boxed{\text{(D) }\sin^{-1}\left(\dfrac{e^{x}+1}{2}\right)+c} \]