Indefinite Integration & Standard Forms

Antiderivative Definition

If $F^{\prime }(x)=f(x)$ on an open interval $I$, then the general antiderivative is $\int f(x)dx=F(x)+C$, where $C$ is an arbitrary constant of integration.

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1. Standard Integral Table

📊Visual Mathematical Intuition

Family of Antiderivative Translates & Integration by Parts Area

Indefinite Family:$\int f(x)dx = F(x) + C \implies F'(x) = f(x) \;\forall C$

By-Parts Formula:$\int u \, dv = u v - \int v \, du \text{ (Rectangle Area Dissection)}$

Family of indefinite antiderivative curves $F(x) + C$ sharing identical tangent slopes at vertical ordinate $x_0$, alongside geometric rectangle dissection of integration by parts $\int u \, dv = uv - \int v \, du$.

$$\begin{array}{rl}\int x^{n}dx& =\frac{x^{n+1}}{n+1}+C(n\ne -1)\\ \int \frac{1}{x}dx& =\ln |x|+C\\ \int e^{ax}dx& =\frac{1}{a}{e}^{ax}+C,\int a^{x}dx=\frac{a^x}{\ln a}+C\\ \int \sin xdx& =-\cos x+C,\int \cos xdx=\sin x+C\\ \int {\sec }^{2}xdx& =\tan x+C,\int {\csc }^{2}xdx=-\cot x+C\\ \int \sec x\tan xdx& =\sec x+C,\int \csc x\cot xdx=-\csc x+C\\ \int \sec xdx& =\ln |\sec x+\tan x|+C=\ln |\tan (\frac{x}{2}+\frac{\pi }{4})|+C\\ \int \frac{dx}{\sqrt{a^2-x^2}}& =\arcsin \left(\frac{x}{a}\right)+C\\ \int \frac{dx}{a^2+x^2}& =\frac{1}{a}\arctan \left(\frac{x}{a}\right)+C\\ \int \frac{dx}{\sqrt{x^2\pm a^2}}& =\ln |x+\sqrt{x^2\pm a^2}|+C\end{array}$$

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2. Integration by Parts & Exponential Standard Form

$$\int u(x)v^{\prime }(x)dx=u(x)v(x)-\int u^{\prime }(x)v(x)dx$$

The Master Exponential Form

$$\int e^x[f(x)+f^{\prime }(x)]dx=e^{x}f(x)+C$$$$\int [f(x)+x{f}^{\prime }(x)]dx=xf(x)+C$$

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3. Algebraic Symmetry & Quadratic Forms

Type: $\int \frac{x^2\pm 1}{x^4+k{x}^2+1}dx$

Divide numerator and denominator by $x^2$:

$$\int \frac{1\pm \frac{1}{x^2}}{{(x\mp \frac{1}{x})}^2+(k\pm 2)}dx$$

Set $t=x\mp \frac{1}{x}⟹dt=(1\pm \frac{1}{x^2})dx$, reducing instantly to standard form $\int \frac{dt}{t^2+A^2}$.