Step 1: The given integral can be written as: \[ I = \int e^x \left( \frac{x}{\sqrt{1+x^2}} + \frac{1}{(1+x^2)^{\frac{3}{2}}} \right) dx \] Let: \[ f(x) = \frac{x}{\sqrt{1+x^2}} \]
Step 2: Now, calculate the derivative of \( f(x) \): \[ f'(x) = \frac{\sqrt{1+x^2} - \frac{x \cdot x}{\sqrt{1+x^2}}}{1+x^2} = \frac{\sqrt{1+x^2} - \frac{x^2}{\sqrt{1+x^2}}}{1+x^2} \] Simplify the numerator: \[ f'(x) = \frac{\sqrt{1+x^2} - \frac{x^2}{\sqrt{1+x^2}}}{1+x^2} = \frac{1}{(1+x^2)^{\frac{3}{2}}} \] Thus, the integral becomes: \[ I = \int e^x \left( f(x) + f'(x) \right) dx \]
Step 3: Using the standard result: \[ \int e^x \left( f(x) + f'(x) \right) dx = e^x f(x) + C \] Substitute \( f(x) = \frac{x}{\sqrt{1+x^2}} \): \[ I = e^x \frac{x}{\sqrt{1+x^2}} + C \]
Final Answer: \[ \boxed{I = e^x \frac{x}{\sqrt{1+x^2}} + C} \]
Explanation: 1. Splitting the Integral: The given integral is split into terms containing \( \frac{x}{\sqrt{1+x^2}} \) and \( \frac{1}{(1+x^2)^{\frac{3}{2}}} \).
2. Defining \( f(x) \): The function \( f(x) \) is chosen as \( \frac{x}{\sqrt{1+x^2}} \) because its derivative results in the second term, \( \frac{1}{(1+x^2)^{\frac{3}{2}}} \).
3. Applying the Formula: The integral formula for \( \int e^x (f(x) + f'(x)) dx = e^x f(x) + C \) is directly applied.
4. Substitution: Finally, substituting \( f(x) \) into the formula gives the result.
A racing track is built around an elliptical ground whose equation is given by \[ 9x^2 + 16y^2 = 144 \] The width of the track is \(3\) m as shown. Based on the given information answer the following: 
(i) Express \(y\) as a function of \(x\) from the given equation of ellipse.
(ii) Integrate the function obtained in (i) with respect to \(x\).
(iii)(a) Find the area of the region enclosed within the elliptical ground excluding the track using integration.
OR
(iii)(b) Write the coordinates of the points \(P\) and \(Q\) where the outer edge of the track cuts \(x\)-axis and \(y\)-axis in first quadrant and find the area of triangle formed by points \(P,O,Q\).