```markdown If

$$\int \frac{(x^2 + 1) e^x}{(x+1)^2} \, dx = f(x) e^x + c$$

then find

$$\frac{d^3 t}{d x^3}$$

at $x = 1$. ```

1. Turning the integral into a differential equation

Because the result of the integral is given as $f(x)e^x + c,$ differentiating both sides immediately gets us back to the integrand:

$\frac{d}{dx}\bigl(f(x)e^x\bigr)\;=\;(f'(x)+f(x))e^x\;=\;\frac{(x^2+1)e^x}{(x+1)^2}.$

Cancelling the common $e^x$ factor, we arrive at a first-order linear ODE:

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

2. Simplifying the right-hand side

Long division (or a quick algebra trick) shows

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

We name this compact form

$g(x)=1-\frac{2}{x+1}+\frac{2}{(x+1)^2}.$

The differential equation is now

$f'(x)+f(x)=g(x).$

3. Finding $f(x)$ quickly

For such a constant-coefficient, first-order linear ODE the integrating-factor method (factor $e^x$) or a clever guess works very fast:

Guess: take $f(x)=A+\frac{B}{x+1}+\frac{C}{(x+1)^2}.$
Differentiate, add to itself, and equate coefficients to $g(x)$. One finds

$A=1,\;B=-2,\;C=0 \;\Longrightarrow\; f(x)=1-\frac{2}{x+1}.$

Note that any extra constant inside $f$ would merely shift the overall “$+c$” of the original integral, so this form is unique for our purpose.

4. Higher-order derivatives

Since $t(x)$ in the problem statement plays the same role as $f(x)$, we set $t(x)=f(x)$:

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

Now work out derivatives systematically:

1. First derivative
   $t'(x)=\frac{2}{(x+1)^2}.$
2. Second derivative
   $t''(x)=-\frac{4}{(x+1)^3}.$
3. Third derivative
   $t'''(x)=\frac{12}{(x+1)^4}.$

5. Evaluate at $x = 1$

A direct substitution gives

$t'''(1)=\frac{12}{(1+1)^4}=\frac{12}{16}=\frac{3}{4}=0.75.$

That is all the theory and logical chain of steps a student needs to reach the requested third derivative at $x=1$.

What is the question asking?

We first have an integral that has already been written in a neat form:

“If $$\int \frac{(x^2+1)e^x}{(x+1)^2},dx = f(x),e^x + c*,”*

The question then asks for the third derivative of a certain function $t$ at $x = 1$.

To keep it simple, think of $f(x)$ like a recipe hidden inside that integral. Once we know $f(x)$, we can call it $t(x)$ and then perform the usual differentiate-thrice job to find $\frac{d^3t}{dx^3}$ at the point $x = 1$.

How would you do it?

1. Peel off the $e^x$ by remembering the product rule:
   If $F(x) = f(x)e^x$, then $\frac{d}{dx}\bigl( f(x)e^x \bigr) = (f'(x)+f(x))e^x.$
2. Match the peeled-off result with the integrand $\frac{(x^2+1)e^x}{(x+1)^2}$. Cancelling $e^x$ gives a tidy differential equation for $f$.
3. Solve the quick, linear differential equation to get a nice algebraic form of $f(x)$.
4. Differentiate $f(x)$ three times (yes, that sounds scary, but if $f$ is a small rational expression, each step is just a power-rule plus chain-rule).
5. Finally, plug in $x = 1$ to see what the third derivative becomes.

That’s exactly what we’ll guide you through in the detailed explanation!

Step-by-step calculation

1. Convert integral into ODE