Integral of the Form $\int e^x[f(x)+f'(x)]\,dx$

Claim: $\int e^x[f(x)+f'(x)]\,dx=e^xf(x)+c$.

Reasoning: let $t=e^xf(x)$. Differentiating, $\dfrac{dt}{dx}=e^x f'(x)+f(x)e^x=e^x[f(x)+f'(x)]$. By definition of integration, $\int e^x[f(x)+f'(x)]\,dx=t+c=e^xf(x)+c$.

For example, since $\frac{d}{dx}\tan x=\sec^2x$, we immediately get $\int e^x[\tan x+\sec^2x]\,dx=e^x\tan x+c$ — no by-parts computation needed, once $f(x)=\tan x$ is spotted. The skill in applying this pattern is entirely in recognising, and if necessary algebraically producing, the shape "$f(x)$ plus its own derivative":

$\int e^x\,\dfrac{2+\sin2x}{1+\cos2x}\,dx$. Using $\sin2x=2\sin x\cos x$ and $1+\cos2x=2\cos^2x$, the fraction is $\dfrac{1}{\cos^2x}+\dfrac{\sin x\cos x}{\cos^2x}=\sec^2x+\tan x$. Taking $f(x)=\tan x$ (so $f'(x)=\sec^2x$), the pattern matches exactly: $I=e^x\tan x+c$.

$\int e^x\,\dfrac{x+2}{(x+3)^2}\,dx$. Write the numerator as $(x+3)-1$: the fraction becomes $\dfrac{1}{x+3}-\dfrac{1}{(x+3)^2}$. Taking $f(x)=\dfrac{1}{x+3}$, note $f'(x)=-\dfrac{1}{(x+3)^2}$ — an exact match: $I=\dfrac{e^x}{x+3}+c$.

$\int e^{\tan^{-1}x}\,\dfrac{1+x+x^2}{1+x^2}\,dx$. Put $t=\tan^{-1}x$, so $x=\tan t$ and $\dfrac{1}{1+x^2}\,dx=dt$. The integrand becomes $e^t[1+\tan t+\tan^2t]=e^t[\tan t+\sec^2t]$ (using $1+\tan^2t=\sec^2t$) — an $e^t[f+f']$ pattern with $f(t)=\tan t$. So $I=e^t\tan t+c=e^{\tan^{-1}x}\cdot x+c$. …