Which Basic Relations in Integration Should You Know?

When we say that integration is the same as “reverse derivation”, what we mean is that $\int f (x) d x$ is what you have to differentiate in order to get $f (x)$ . So another word for integration is anti-derivation.

Even though integration might seem a bit cryptic in the beginning, the sums and differences of functions, and products and quotients of functions and numbers, all follow some simple rules. These rules show you how to integrate in these instances:

Rule

Useful Integration Rules

\begin{array}{l} \int u (x) + v (x) d x & = \int u (x) d x + \int v (x) d x \\ \int u (x) - v (x) d x & = \int u (x) d x - \int v (x) d x \end{array}

\begin{array}{l} \int k u (x) d x & = k \int u (x) d x \\ \int \frac{u (x)}{k} d x & = \frac{1}{k} \int u (x) d x \end{array}

\begin{array}{l} \int u (x) + v (x) d x & = \int u (x) d x + \int v (x) d x \\ \int u (x) - v (x) d x & = \int u (x) d x - \int v (x) d x \\ \int k u (x) d x & = k \int u (x) d x \\ \int \frac{u (x)}{k} d x & = \frac{1}{k} \int u (x) d x \end{array}

Example 1

Solve the integral $\int 2 \cos (2 x) + \frac{1}{x} d x$

$\begin{array}{l} \int 2 \cos (2 x) + \frac{1}{x} d x \\ = 2 \cdot \frac{1}{2} \sin (2 x) + \ln | x | + C \\ = \sin (2 x) + \ln | x | + C \end{array}$

$\begin{array}{l} \int 2 \cos (2 x) + \frac{1}{x} d x & = 2 \cdot \frac{1}{2} \sin (2 x) + \ln | x | + C \\ = \sin (2 x) + \ln | x | + C \end{array}$

Example 2

Solve the integral $\int e^{3 x} - \sin (π x) + π d x$

$\begin{array}{l} \int e^{3 x} - \sin (π x) + π d x \\ = \frac{1}{3} e^{3 x} + \frac{1}{π} \cos (π x) + π x + C \end{array}$

$\int e^{3 x} - \sin (π x) + π d x = \frac{1}{3} e^{3 x} + \frac{1}{π} \cos (π x) + π x + C$

Example 3

Solve the integral $\int 3 x^{4} + 3 \tan (4 x) d x$

$\begin{array}{l} \int 3 x^{4} + 3 \tan (4 x) d x \\ = 3 \frac{1}{5} x^{5} - 3 \cdot \frac{1}{4} \ln | \cos (4 x) | + C \\ = \frac{3}{5} x^{5} - \frac{3}{4} \ln | \cos (4 x) | + C \end{array}$

$\begin{array}{l} \int 3 x^{4} + 3 \tan (4 x) d x & = 3 \frac{1}{5} x^{5} - 3 \cdot \frac{1}{4} \ln | \cos (4 x) | + C \\ = \frac{3}{5} x^{5} - \frac{3}{4} \ln | \cos (4 x) | + C \end{array}$

Example 4

Solve the integral $\int 2^{x} - \frac{5}{3 x + 4} d x$

$\begin{matrix} \int 2^{x} - \frac{5}{3 x + 4} d x \\ = \frac{2^{x}}{\ln 2} - 5 \cdot \frac{1}{3} \ln | 3 x + 4 | + C \\ = \frac{2^{x}}{\ln 2} - \frac{5}{3} \ln | 3 x + 4 | + C \end{matrix}$

$\begin{array}{l} \int 2^{x} - \frac{5}{3 x + 4} d x & = \frac{2^{x}}{\ln 2} - 5 \cdot \frac{1}{3} \ln | 3 x + 4 | + C \\ = \frac{2^{x}}{\ln 2} - \frac{5}{3} \ln | 3 x + 4 | + C \end{array}$

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