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

Question

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

Accepted Answer

Write $x^2+1 = (x^2-1)+2 = (x-1)(x+1)+2$, so

$$\frac{(x^2+1)e^x}{(x+1)^2} = e^x\left[\frac{x-1}{x+1}+\frac{2}{(x+1)^2}\right]$$

Recognizing this as $e^x[h(x)+h'(x)]$ with $h(x)=\dfrac{x-1}{x+1}$:

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

$$f'(x) = \frac2{(x+1)^2},\quad f''(x)=-\frac4{(x+1)^3},\quad f'''(x)=\frac{12}{(x+1)^4}$$

$$f'''(1) = \frac{12}{16} = \frac34$$

Solutions · 1