Example 22: $\int e^x\big[f(x)+f'(x)\big]dx$⁠: two integrals

Basics you need

Part (i): $\int e^x\left(\tan^{-1}x + \frac{1}{1+x^2}\right)dx$

1. Why $\int e^x[f + f']\,dx = e^x f + C$

   Let $f$ be any differentiable function. By the product rule,

   $$\begin{aligned} \frac{d}{dx}\big[e^x f(x)\big] &= e^x f(x) + e^x f'(x) \\ &= e^x\big[f(x) + f'(x)\big]. \end{aligned}$$

   So $e^x f(x)$ is an antiderivative of $e^x[f(x) + f'(x)]$⁠, and

   $$\int e^x\big[f(x) + f'(x)\big]\,dx = e^x f(x) + C.$$
   The book proves the same result by integration by parts: it splits the integral as $\int e^x f\,dx + \int e^x f'\,dx$⁠, integrates the first by parts with $f$ first and $e^x$ second, getting $f e^x - \int f' e^x\,dx$⁠, and the two integrals $\mp\int e^x f'\,dx$ cancel. The product rule route above is the same fact read the other way.
   Product rule; $\frac{d}{dx}e^x = e^x$⁠. Lesson: integration by parts

2. Spot $f$ and write the answer

   Take $f(x) = \tan^{-1}x$⁠. Then $f'(x) = \frac{1}{1+x^2}$⁠, so the integrand is exactly $e^x[f(x) + f'(x)]$⁠. Hence

   $$\int e^x\left(\tan^{-1}x + \frac{1}{1+x^2}\right)dx = e^x\tan^{-1}x + C.$$
   Of the two terms in the bracket, one must be the derivative of the other. $\frac{1}{1+x^2}$ is the derivative of $\tan^{-1}x$⁠, and not the other way round, so $f = \tan^{-1}x$⁠.
   $\frac{d}{dx}\tan^{-1}x = \frac{1}{1+x^2}$ (formula (x)). Basics: trigonometric identities

3. Check: differentiate the answer

   $$\frac{d}{dx}\big(e^x\tan^{-1}x\big) = e^x\tan^{-1}x + e^x\cdot\frac{1}{1+x^2},$$

   which is $e^x\left(\tan^{-1}x + \frac{1}{1+x^2}\right)$⁠, the integrand, for every real $x$⁠.

   The product rule produces exactly the two terms of the bracket.
   Product rule. Basics: integration undoes differentiation

Part (ii): $\int\frac{(x^2+1)\,e^x}{(x+1)^2}\,dx$

1. Rewrite the fraction as "a function plus its derivative"

   Write $x^2 + 1 = (x^2 - 1) + 2$⁠. Then, for $x \ne -1$⁠,

   $$\begin{aligned} \frac{x^2+1}{(x+1)^2} &= \frac{x^2-1}{(x+1)^2} + \frac{2}{(x+1)^2} \\ &= \frac{(x-1)(x+1)}{(x+1)^2} + \frac{2}{(x+1)^2} \\ &= \frac{x-1}{x+1} + \frac{2}{(x+1)^2}. \end{aligned}$$

   So the integral is $\int e^x\left[\frac{x-1}{x+1} + \frac{2}{(x+1)^2}\right]dx$⁠.

   We want the bracket to be $f + f'$⁠. Adding and subtracting 1 makes $x^2 - 1 = (x-1)(x+1)$ appear, which cancels one factor $x + 1$⁠. The leftover $\frac{2}{(x+1)^2}$ has a squared bottom, which is what a derivative of a fraction with bottom $x + 1$ looks like. Cancelling $x + 1$ is allowed because $x \ne -1$⁠.
   $a^2 - b^2 = (a-b)(a+b)$⁠. Basics: splitting fractions

2. Spot $f$ and write the answer

   Take $f(x) = \frac{x-1}{x+1}$⁠. By the quotient rule,

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

   So the bracket is $f(x) + f'(x)$⁠, and

   $$\int\frac{(x^2+1)\,e^x}{(x+1)^2}\,dx = \frac{x-1}{x+1}\,e^x + C.$$
   The rule of Part (i), Step 1 applies on any interval where $f$ is differentiable, here on $(-\infty, -1)$ and on $(-1, \infty)$⁠.
   Quotient rule (Chapter 5). Lesson: integration by parts

3. Check: differentiate the answer

   $$\begin{aligned} \frac{d}{dx}\left[\frac{x-1}{x+1}\,e^x\right] &= \frac{x-1}{x+1}\,e^x + \frac{2}{(x+1)^2}\,e^x \\ &= e^x\,\frac{(x-1)(x+1) + 2}{(x+1)^2} \\ &= \frac{(x^2+1)\,e^x}{(x+1)^2}. \end{aligned}$$
   $(x-1)(x+1) + 2 = x^2 - 1 + 2 = x^2 + 1$⁠, so the derivative is exactly the integrand for every $x \ne -1$⁠.
   Product rule and the derivative found in Part (ii), Step 2. Basics: integration undoes differentiation

Every fact used, and where it comes from