Properties of Indefinite Integrals and Direct Applications

Three simple but powerful properties let us break a complicated integrand apart and integrate it piece by piece.

Theorem 1 (sum rule). If $f$ and $g$ are integrable functions of $x$, then $\int[f(x)+g(x)]\,dx=\int f(x)\,dx+\int g(x)\,dx$.

Reasoning: let $\int f(x)\,dx=g_1(x)+c_1$ and $\int g(x)\,dx=g_2(x)+c_2$. By the chain rule, $\frac{d}{dx}\big[(g_1(x)+c_1)+(g_2(x)+c_2)\big]=\frac{d}{dx}(g_1(x)+c_1)+\frac{d}{dx}(g_2(x)+c_2)=f(x)+g(x)$. Since the derivative of $(g_1+c_1)+(g_2+c_2)$ is exactly $f(x)+g(x)$, this combined expression is by definition a primitive of $f(x)+g(x)$, which is exactly the claimed sum rule.

Theorem 2 (difference rule). $\int[f(x)-g(x)]\,dx=\int f(x)\,dx-\int g(x)\,dx$ — proved by the identical argument, replacing every $+$ with $-$.

Theorem 3 (constant-multiple rule). For a constant $k$, $\int k\,f(x)\,dx=k\int f(x)\,dx$ — proved the same way, since $\frac{d}{dx}\big[k(g_1(x)+c_1)\big]=k\,f(x)$.

Together these three rules mean an integral can be split term by term, with constant coefficients carried straight through, and each piece matched against the elementary formulae of 3.1.1. A few representative applications:

$\int(x^3+3x)\,dx$. Split by the sum rule: $\int x^3\,dx+\int 3x\,dx = \frac{x^4}{4}+\frac{3^x}{\log 3}+c$.

$\int\left(\sin x+\frac1x+\frac{1}{\sqrt[3]{x}}\right)dx$. Term by term, using $x^{-1/3}$ for the cube-root term: $-\cos x+\log x+\frac{x^{2/3}}{2/3}+c=-\cos x+\log x+\frac32x^{2/3}+c$.

$\int(\tan x+\cot x)^2\,dx$. Expand: $\tan^2x+2\tan x\cot x+\cot^2x = \tan^2x+2+\cot^2x$ (since $\tan x\cot x=1$). Rewrite $\tan^2x=\sec^2x-1$ and $\cot^2x=\csc^2x-1$: the expression collapses to $\sec^2x+\csc^2x$, so the integral is $\tan x-\cot x+c$.

$\int\frac{\sqrt x+1}{x+\sqrt x}\,dx$. Factor the denominator as $\sqrt x(\sqrt x+1)$; the $(\sqrt x+1)$ cancels, leaving $\int\frac{1}{\sqrt x}\,dx=2\sqrt x+c$.

$\int\frac{e^{4\log x}-e^{5\log x}}{x^5}\,dx$. Using $e^{k\log x}=x^k$, the numerator is $x^4-x^5$, so the integrand simplifies to $\frac1x-1$, giving $\log x-x+c$.

$\int\frac{2x+3}{5x-1}\,dx$. An improper algebraic fraction — divide first: $2x+3=\frac25(5x-1)+\frac{17}{5}$, so the integrand is $\frac25+\frac{17/5}{5x-1}$, giving $\frac25x+\frac{17}{25}\log(5x-1)+c$.

$\int\frac{1}{\sqrt{3x+1}-\sqrt{3x-5}}\,dx$. Rationalise by multiplying top and bottom by the conjugate $\sqrt{3x+1}+\sqrt{3x-5}$; the denominator becomes $(3x+1)-(3x-5)=6$, leaving $\frac16\int\left[(3x+1)^{1/2}+(3x-5)^{1/2}\right]dx=\frac{1}{27}\left[(3x+1)^{3/2}+(3x-5)^{3/2}\right]+c$. …