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Mathematics Question on applications of integrals

if ∫(x2+1)ex(x+1)2dx=ƒ(x)ex+C∫ \frac{(x^2+1)e^x}{(x+1)^2}dx = ƒ(x)e^x+C where CC is a constant, then d3ƒdx3\frac{d^3ƒ}{dx^3} at x=1x = 1 is equal to :

A

−34-\frac{3}{4}

B

34\frac{3}{4}

C

−32-\frac{3}{2}

D

32\frac{3}{2}

Answer

34\frac{3}{4}

Explanation

Solution

I=∫(x2+1)ex(x+1)2dx=ƒ(x)ex+CI = ∫ \frac{(x^2+1)e^x}{(x+1)^2}dx = ƒ(x)e^x+C

I=∫ex(x2−1+1+1)(x+1)2dxI = ∫ \frac{e^x(x^2-1+1+1)}{(x+1)^2}dx

= ∫ex[x−1x+1+2(x+1)2]dx∫ e^x\bigg[\frac{x-1}{x+1}+\frac{2}{(x+1)^2}\bigg]dx

= ex(x−1x+1)+ce^x\bigg(\frac{x-1}{x+1}\bigg)+c

∴f(x)=x−1x+1∴ f (x) = \frac{x-1}{x+1}

f(x)=1−2x+1f(x) = \frac{1- 2}{x+1}

f′(x)=2(1x+1)2f' (x) = 2\bigg(\frac{1}{x+1}\bigg)^2

f′′(x)=−4(1x+1)3f''(x) = -4\bigg(\frac{1}{x+1}\bigg)^3

f′′′(x)=12(x+1)4f'''(x) = \frac{12}{(x+1)^4}

for x=1x = 1

f′′′(1)=1224f'''(1) = \frac{12}{24}

= 1216\frac{12}{16}

= 34\frac{3}{4}

Hence, the correct option is (B): 34\frac{3}{4}