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Step 1: Express the Integral in a Suitable Form
We start with the integral:
$I = \int \frac{(x^2 + 1)e^x}{(x+1)^2} \, dx$
The given statement says this integral equals $f(x)e^x + C$. Our aim is to find $f(x)$, then compute $f'''(x)$ and finally evaluate it at $x = 1$.
Step 2: Simplify the Integrand
Rewrite $x^2 + 1$ as $(x^2 - 1) + 2$, so the integrand becomes:
$\displaystyle \int \frac{(x^2 - 1 + 2)e^x}{(x+1)^2} \, dx
\;=\; \int e^x \Bigl[\frac{x^2 - 1}{(x+1)^2} + \frac{2}{(x+1)^2}\Bigr] \, dx.$
Notice that $\frac{x^2 - 1}{(x+1)^2}$ can be factored or rewritten in a simpler form to facilitate integration.
Step 3: Split and Rewrite the Terms
Observe that $x^2 - 1 = (x - 1)(x + 1)$. Hence,
$\frac{x^2 - 1}{(x+1)^2} = \frac{(x - 1)(x+1)}{(x+1)^2} = \frac{x - 1}{x+1}.$
So the integrand becomes:
$ \int e^x \bigl[\frac{x - 1}{x+1} + \frac{2}{(x+1)^2}\bigr] \, dx.$
Step 4: Identify the Result of the Integral
It can be shown (by the method of differentiation of a product $f(x)e^x$ or by manual integration) that the antiderivative simplifies to:
$I = \int \frac{(x^2 + 1)e^x}{(x+1)^2} \, dx = f(x)e^x + C,$
where
$f(x) = \frac{x - 1}{x+1} \,=\, 1 - \frac{2}{x+1}.
Step 5: Compute the First and Second Derivatives of f(x)
$f(x) = 1 - \frac{2}{x+1}.$
$f'(x) = -2 \cdot \Bigl(-\frac{1}{(x+1)^2}\Bigr) = \frac{2}{(x+1)^2}.
$f''(x) = 2 \cdot \frac{d}{dx}\Bigl(\frac{1}{(x+1)^2}\Bigr)
= 2 \cdot \Bigl(-\frac{2}{(x+1)^3}\Bigr)
= -\frac{4}{(x+1)^3}.
Step 6: Compute the Third Derivative of f(x)
Differentiate $f''(x)$ once more:
$f'''(x) = -4 \cdot \frac{d}{dx}\Bigl(\frac{1}{(x+1)^3}\Bigr)
= -4 \cdot \Bigl(-\frac{3}{(x+1)^4}\Bigr)
= \frac{12}{(x+1)^4}.$
Step 7: Evaluate $f'''(x)$ at $x=1$
$f'''(1) = \frac{12}{(1+1)^4}
= \frac{12}{2^4}
= \frac{12}{16}
= \frac{3}{4}.
Hence, $\frac{d^3 f}{dx^3}$ at $x = 1$ is $\displaystyle \frac{3}{4}.$
