The direct integrals are the simplest type of integral that exists, since they are solved straight away by applying what we know, which is essentially that integrating is the opposite of deriving.
These are the integrals of functions of the type $$k \cdot x^n$$, where $$k$$ is a constant and $$n$$ is any number other than $$-1$$.
We have then :$$$\displaystyle \int k \cdot x^n \ dx= k \cdot \dfrac{x^{n+1}}{n+1}+C$$$ note that if we take the derivative we obtain: $$$\dfrac{d}{dx}\Big(k \cdot \dfrac{x^{n+1}}{n+1}+C\Big)=\dfrac{d}{dx}\Big(k \cdot \dfrac{x^{n+1}}{n+1}\Big)+\dfrac{d}{dx}C= k\cdot \dfrac{d}{dx}\Big(\dfrac{x^{n+1}}{n+1}\Big)=$$$ $$$= k \cdot \dfrac{(n+1)x^n}{n+1}=k \cdot x^n$$$
$$\displaystyle \int x \ dx= \dfrac{x^2}{2}+C$$, since $$\dfrac{d}{dx}\Big(\dfrac{x^2}{2}+C\Big)=x$$
$$\displaystyle \int x^2 \ dx= \dfrac{x^3}{3}+C$$, since $$\dfrac{d}{dx}\Big(\dfrac{x^3}{3}+C\Big)=x^2$$
$$\displaystyle \int x^3 \ dx= \dfrac{x^4}{4}+C$$, since $$\dfrac{d}{dx}\Big(\dfrac{x^4}{4}+C\Big)=x^3$$
$$\displaystyle \int 23x^5 \ dx= 23\dfrac{x^6}{6}+C$$, since $$\dfrac{d}{dx}\Big(23\dfrac{x^6}{6}+C\Big)=23 \cdot \dfrac{d}{dx}\Big(\dfrac{x^6}{6}\Big)=23 \cdot 6 \cdot \dfrac{x^{6-1}}{6}=23 \cdot x^5$$
$$\displaystyle \int \sqrt{x} \ dx = \int x^{\frac{1}{2}}=\dfrac{x^{\frac{1}{2}+1}}{\dfrac{1}{2}+1}+C=\dfrac{x^\frac{3}{2}}{\dfrac{3}{2}}+C=\dfrac{2}{3}x^\frac{3}{2}$$, since $$\dfrac{d}{dx}\Big(\dfrac{2}{3}x^\frac{3}{2}+C\Big)=\dfrac{d}{dx}\Big(\dfrac{2}{3}x^\frac{3}{2}\Big)+\dfrac{d}{dx}C=\dfrac{2}{3}\cdot \dfrac{d}{dx}x^\frac{2}{3}=\dfrac{2}{3}\cdot \dfrac{3}{2}x^\frac{1}{2}=x^\frac{1}{2}=\sqrt{x}$$
$$\displaystyle \int \sqrt[3]{425} \cdot x^\frac{2}{3} \ dx= \sqrt[3]{425} \cdot \dfrac{x^\frac{5}{3}}{\dfrac{5}{3}}+C$$, since $$\dfrac{d}{dx}\Big(\sqrt[3]{425} \cdot \dfrac{x^\frac{5}{3}}{\dfrac{5}{3}}+C\Big)=\dfrac{d}{dx}\Big(\sqrt[3]{425} \cdot \dfrac{x^\frac{5}{3}}{\dfrac{5}{3}}\Big)+\dfrac{d}{dx}\Big(C\Big)=\sqrt[3]{425} \cdot \dfrac{d}{dx}\Big( \dfrac{x^\frac{5}{3}}{\dfrac{5}{3}}\Big)=$$
$$=\sqrt[3]{425} \cdot x^\frac{2}{3}$$
$$\displaystyle \int 15x^{-2} \ dx= 15\dfrac{x^{-1}}{-1}+C=-15\dfrac{1}{x}+C$$, since $$\dfrac{d}{dx}\Big(-15\dfrac{1}{x}\Big)=-15 \cdot \dfrac{d}{dx}\Big(\dfrac{1}{x}\Big)=-15 \cdot \dfrac{x^{-2}}{-1}=15 \cdot x^{-2}$$