∫▒ ((x^2+1) e^x)/((x+1)^2 ) dx=f(x)e^x+C, where C is a constant, then (d^3 f)/(dx^3 ) at x=1 is скачать в хорошем качестве

∫▒ ((x^2+1) e^x)/((x+1)^2 ) dx=f(x)e^x+C, where C is a constant, then (d^3 f)/(dx^3 ) at x=1 is

∫▒ ((x^2+1) e^x)/((x+1)^2 ) dx=f(x)e^x+C, where C is a constant, then (d^3 f)/(dx^3 ) at x=1 is equal to Description: The given mathematical problem involves evaluating the integral: \[ \int \frac{(x^2 + 1)e^x}{(x+1)^2} dx = f(x)e^x + C \] where \( C \) is a constant. The task is to determine the third derivative \( \frac{d^3 f}{dx^3} \) at \( x = 1 \). Hash Tags: #Calculus, #Integration, #Derivatives, #Mathematics, #JEE, #MathProblem, #Differentiation, #higherorderderivatives We are given the integral: \[ \int \frac{(x^2 + 1)e^x}{(x+1)^2} dx = f(x)e^x + C \] Step 1: Identify \( f(x) \) Comparing both sides, we differentiate both sides w.r.t. \( x \): \[ \frac{(x^2 + 1)e^x}{(x+1)^2} = \left( f(x) e^x \right)' \] Using the product rule: \[ \frac{(x^2 + 1)e^x}{(x+1)^2} = f(x) e^x + f'(x) e^x \] Dividing by \( e^x \) throughout: \[ \frac{x^2 + 1}{(x+1)^2} = f(x) + f'(x) \] Thus, we get the differential equation: \[ f'(x) + f(x) = \frac{x^2 + 1}{(x+1)^2} \] Step 2: Solve for \( f(x) \) This is a first-order linear differential equation. The integrating factor (IF) is: \[ I.F. = e^{\int 1 dx} = e^x \] Multiplying throughout by \( e^x \): \[ e^x f'(x) + e^x f(x) = \frac{(x^2 + 1)e^x}{(x+1)^2} \] Rewriting: \[ \frac{d}{dx} \left( f(x) e^x \right) = \frac{(x^2 + 1)e^x}{(x+1)^2} \] Integrating both sides: \[ f(x) e^x = \int \frac{(x^2 + 1)e^x}{(x+1)^2} dx \] Using substitution \( t = x + 1 \), so that \( dt = dx \), we get: \[ \int \frac{(t^2 - 2t + 2)e^x}{t^2} dx \] Splitting terms: \[ \int \left( 1 - \frac{2t - 2}{t^2} \right) e^x dx \] Solving term-wise: \[ \int e^x dx - 2 \int \frac{(t-1)e^x}{t^2} dx \] Using standard integral forms, solving, and simplifying, we get: \[ f(x) = \frac{x}{x+1} \] Step 3: Compute \( f'''(x) \) at \( x = 1 \) We differentiate \( f(x) \): \[ f(x) = \frac{x}{x+1} \] #### First derivative: Using quotient rule: \[ f'(x) = \frac{(x+1) \cdot 1 - x \cdot 1}{(x+1)^2} = \frac{x+1 - x}{(x+1)^2} = \frac{1}{(x+1)^2} \] #### Second derivative: \[ f''(x) = \frac{d}{dx} \left( (x+1)^{-2} \right) = -2(x+1)^{-3} \] #### Third derivative: \[ f'''(x) = -2 \cdot (-3) (x+1)^{-4} = 6(x+1)^{-4} \] Substituting \( x = 1 \): \[ f'''(1) = 6(1+1)^{-4} = 6 \times \frac{1}{16} = \frac{6}{16} = \frac{3}{8} \] Final Answer: \[ \frac{3}{8} \] JOIN LIVE CLASSES https://chat.whatsapp.com/GmV7W51d8LT... website : www.e-math.in contact : +91 7032507699(whatsapp) visit my website: www.e-math.in youtubechannel:    / onlinemathstutionguntur   facebook page:   / emath2016   twitter :  / e-math2016   whatsapp :+91 7032507699 e-mail : [email protected]