Question

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

• A. $-\frac{3}{4}$
• B. $\frac{3}{4}$
• C. $-\frac{3}{2}$
• D. $\frac{3}{2}$

Answer 1

The Correct Option is B

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

$I = ∫ \frac{e^x(x^2-1+1+1)}{(x+1)^2}dx$

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

= $e^x\bigg(\frac{x-1}{x+1}\bigg)+c$

$∴ f (x) = \frac{x-1}{x+1}$

$f(x) = \frac{1- 2}{x+1}$

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

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

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

for $x = 1$

$f'''(1) = \frac{12}{24}$

= $\frac{12}{16}$

= $\frac{3}{4}$

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