Solveeit Logo
Login

Question

Question: If $\int \frac{(x^{2}+1)e^{x}}{(x+1)^{2}}dx=f(x)e^{x}+C$, where C is a constant, then $\frac{d^{3}f}...

If ∫(x2+1)ex(x+1)2dx=f(x)ex+C\int \frac{(x^{2}+1)e^{x}}{(x+1)^{2}}dx=f(x)e^{x}+C, where C is a constant, then d3fdx3\frac{d^{3}f}{dx^{3}} at x=1x=1 is equal to:

Answer

3/4

Explanation

Solution

To solve the given problem, we first need to determine the function f(x)f(x) by evaluating the integral.

The given integral is ∫(x2+1)ex(x+1)2dx\int \frac{(x^{2}+1)e^{x}}{(x+1)^{2}}dx. We can rewrite the numerator x2+1x^2+1 in terms of (x+1)(x+1): x2+1=x2−1+2=(x−1)(x+1)+2x^2+1 = x^2-1+2 = (x-1)(x+1)+2.

Substitute this into the integrand: (x2+1)ex(x+1)2=((x−1)(x+1)+2)ex(x+1)2=((x−1)(x+1)(x+1)2+2(x+1)2)ex\frac{(x^{2}+1)e^{x}}{(x+1)^{2}} = \frac{((x-1)(x+1)+2)e^{x}}{(x+1)^{2}} = \left(\frac{(x-1)(x+1)}{(x+1)^{2}} + \frac{2}{(x+1)^{2}}\right)e^{x} =(x−1x+1+2(x+1)2)ex= \left(\frac{x-1}{x+1} + \frac{2}{(x+1)^{2}}\right)e^{x}

This integral is of the form ∫(g(x)+g′(x))exdx=g(x)ex+C\int (g(x) + g'(x))e^x dx = g(x)e^x + C. Let's identify g(x)g(x). Let g(x)=x−1x+1g(x) = \frac{x-1}{x+1}. Now, we find the derivative of g(x)g(x): g′(x)=ddx(x−1x+1)=(1)(x+1)−(x−1)(1)(x+1)2=x+1−x+1(x+1)2=2(x+1)2g'(x) = \frac{d}{dx}\left(\frac{x-1}{x+1}\right) = \frac{(1)(x+1) - (x-1)(1)}{(x+1)^2} = \frac{x+1-x+1}{(x+1)^2} = \frac{2}{(x+1)^2} As we can see, the integrand is indeed in the form (g(x)+g′(x))ex(g(x) + g'(x))e^x.

Therefore, the integral is: ∫(x−1x+1+2(x+1)2)exdx=x−1x+1ex+C\int \left(\frac{x-1}{x+1} + \frac{2}{(x+1)^{2}}\right)e^{x} dx = \frac{x-1}{x+1}e^{x} + C

Comparing this with the given equation ∫(x2+1)ex(x+1)2dx=f(x)ex+C\int \frac{(x^{2}+1)e^{x}}{(x+1)^{2}}dx=f(x)e^{x}+C, we can identify f(x)f(x): f(x)=x−1x+1f(x) = \frac{x-1}{x+1}

Now, we need to find the third derivative of f(x)f(x), i.e., d3fdx3\frac{d^{3}f}{dx^{3}}, and then evaluate it at x=1x=1.

First derivative: f′(x)=ddx(x−1x+1)=2(x+1)2f'(x) = \frac{d}{dx}\left(\frac{x-1}{x+1}\right) = \frac{2}{(x+1)^2} (This was already calculated as g′(x)g'(x)).

Second derivative: f′′(x)=ddx(2(x+1)2)=2ddx((x+1)−2)f''(x) = \frac{d}{dx}\left(\frac{2}{(x+1)^2}\right) = 2 \frac{d}{dx}((x+1)^{-2}) f′′(x)=2(−2)(x+1)−3(1)=−4(x+1)−3=−4(x+1)3f''(x) = 2(-2)(x+1)^{-3}(1) = -4(x+1)^{-3} = \frac{-4}{(x+1)^3}

Third derivative: f′′′(x)=ddx(−4(x+1)3)=−4ddx((x+1)−3)f'''(x) = \frac{d}{dx}\left(\frac{-4}{(x+1)^3}\right) = -4 \frac{d}{dx}((x+1)^{-3}) f′′′(x)=−4(−3)(x+1)−4(1)=12(x+1)−4=12(x+1)4f'''(x) = -4(-3)(x+1)^{-4}(1) = 12(x+1)^{-4} = \frac{12}{(x+1)^4}

Finally, evaluate f′′′(x)f'''(x) at x=1x=1: f′′′(1)=12(1+1)4=1224=1216f'''(1) = \frac{12}{(1+1)^4} = \frac{12}{2^4} = \frac{12}{16} Simplify the fraction: f′′′(1)=3×44×4=34f'''(1) = \frac{3 \times 4}{4 \times 4} = \frac{3}{4}