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 stationary tank is cylindrical in shape with two hemispherical ends and is horizontal, as shown in the figure. \(R\) is the radius of the cylinder as well as of the hemispherical ends. The tank is half filled with an oil of density \(\rho\) and the rest of the space in the tank is occupied by air. The air pressure, inside the tank as well as outside it, is atmospheric. The acceleration due to gravity (\(g\)) acts vertically downward. The net horizontal force applied by the oil on the right hemispherical end (shown by the bold outline in the figure) is:
Draw a rough sketch for the curve $y = 2 + |x + 1|$. Using integration, find the area of the region bounded by the curve $y = 2 + |x + 1|$, $x = -4$, $x = 3$, and $y = 0$.
Let \[ I(x) = \int \frac{dx}{(x-11)^{\frac{11}{13}} (x+15)^{\frac{15}{13}}} \] If \[ I(37) - I(24) = \frac{1}{4} \left( b^{\frac{1}{13}} - c^{\frac{1}{13}} \right) \] where \( b, c \in \mathbb{N} \), then \[ 3(b + c) \] is equal to: