Logarithmic, exponential and trigonometric direct integrals

Direct integrals for the logarithm

These are the integrals of functions of the type $k \cdot x^{- 1}$ and they are solved as follows:

$\int k \frac{1}{x} d x = k \int x^{- 1} d x = k \ln | x | + C$

since

$\frac{d}{d x} (k \ln | x | + C) = \frac{d}{d x} k \ln | x | + 0 = k \frac{d}{d x} \ln | x | = k \frac{1}{x}$

Let's observe that the result is the logarithm of the absolute value, for the logarithm of a negative number does not exist.

Due to the properties of the logarithms, if $a > 0$ :

$\int \frac{1}{x \cdot \ln a} d x = \log_{a} | x | + C$

Example

$\int \frac{1}{x} d x = \ln | x | + C$

$\int 15 \cdot \frac{1}{x} d x = 15 \int \frac{1}{x} d x = 15 \ln | x | + C$

$\int \frac{1}{x \ln 5} d x = \log_{5} | x | + C$

Since we know that the derivative of the function $e^{x}$ is itself then,

$\int e^{x} d x = e^{x} + C$

and using the properties of the logarithms, if $a > 0$ and $a \neq 1$ :

$\int a^{x} d x = \frac{a^{x}}{l n a} + C$

Example

$\int 3 e^{x} d x = 3 e^{x} + C$

$\int 3^{x} d x = \int 3^{x} d x = \frac{3^{x}}{\ln 3} + C$

$\int 4^{x} + 4 e^{x} d x = \int 4^{x} d x + 4 \int e^{x} d x = \frac{4^{x}}{l n 4} + 4 e^{x} + C$

We can use what we learned about taking the derivatives of trigonometric functions (sine, cosine, tangent, arctangent, etc.) to integrate:

$\int \sin x d x = - \cos x + C$

$\int \cos x d x = \sin x + C$

$\int \frac{1}{\cos^{2} x} d x = \tan x + C$

$\int \frac{1}{\sqrt{1 - x^{2}}} d x = \arcsin x + C$

$\int \frac{- 1}{\sqrt{1 - x^{2}}} d x = \arccos x + C$

$\int \frac{1}{a + x^{2}} d x = \arctan x + C$

Example

$\int 2 \sin x d x = - 2 \cos x + C$

$\int 5 \cos x + 3 \sin x d x = 5 \int \cos x d x + 3 \int \sin x d x = 5 \sin x - 3 \cos x + C$

$\int \frac{3}{1 + x^{2}} d x = 3 \arctan x + C$