Direct integrals for the logarithm
These are the integrals of functions of the type $$k\cdot x^{-1} $$ and they are solved as follows:
$$$\displaystyle \int k\frac{1}{x} \ dx= k \int x^{-1} \ dx = k\ln |x| +C$$$
since
$$$ \frac{d}{dx}\Big( k\ln |x|+C\Big)=\frac{d}{dx} k\ln |x|+0= k\frac{d}{dx}\ln |x|= k \frac{1}{x}$$$
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 $$a> 0$$:
$$$\displaystyle \int \frac{1}{x \cdot \ln a} \ dx= \log_{a}|x|+C$$$
$$$\displaystyle \int \frac{1}{x} \ dx=\ln |x| +C$$$
$$$\displaystyle \int 15 \cdot \frac{1}{x} \ dx = 15 \int \frac{1}{x} \ dx= 15 \ln |x|+C$$$
$$$\displaystyle \int \frac{1}{x\ln 5} \ dx= \log_{5} |x| +C$$$
Exponential integrals
Since we know that the derivative of the function $$e^x$$ is itself then,
$$$\displaystyle \int e^x \ dx = e^x+C$$$
and using the properties of the logarithms, if $$a> 0$$ and $$a\neq 1$$:
$$$\displaystyle \int a^x \ dx= \frac{a^x}{ln a}+C$$$
$$$\displaystyle \int 3e^x \ dx= 3e^x+C$$$
$$$\displaystyle \int 3^x \ dx = \int 3^x \ dx = \frac{3^x}{\ln 3} +C$$$
$$$\displaystyle \int 4^x+4e^x \ dx =\int 4^x \ dx + 4 \int e^x \ dx= \frac{4^x}{ln 4}+4e^x+C$$$
Trigonometric integrals
We can use what we learned about taking the derivatives of trigonometric functions (sine, cosine, tangent, arctangent, etc.) to integrate:
$$$\displaystyle \int \sin x \ dx = - \cos x +C$$$
$$$\displaystyle \int \cos x \ dx= \sin x +C$$$
$$$\displaystyle \int \frac{1}{\cos^2 x} \ dx = \tan x +C$$$
$$$\displaystyle \int \frac{1}{\sqrt{1-x^2}} \ dx= \arcsin x+C$$$
$$$\displaystyle \int \frac{-1}{\sqrt{1-x^2}} \ dx= \arccos x +C$$$
$$$\displaystyle \int \frac{1}{a+x^2} \ dx = \arctan x +C$$$
$$$\displaystyle \int 2 \sin x \ dx= -2 \cos x +C$$$
$$$\displaystyle \int 5 \cos x+ 3 \sin x \ dx = 5 \int \cos x \ dx + 3 \int \sin x \ dx= 5 \sin x -3 \cos x +C$$$
$$$\displaystyle \int \frac{3}{1+x^2} \ dx = 3 \arctan x + C$$$