Take an example:

find anti-derivative $F(x)$ of function

$f(x)=5 x^4+2 \text { s.t. } F(1)=5$

$\frac{d}{d x}(x^5+2 x)=5x^4+2$

$F(x)=x^5+2 x+c$

$F(1)=5 \Rightarrow 1+2+c=5 \Rightarrow c=2$

$F(x)=x^5+2 x+2$

1. a) $\frac{d}{d x} \int f(x) d x=f(x)$

b) $\int f^{\prime}(x) d x=f(x)+c, c \in R$.

Proof:

(a) $~ \frac{d}{d x}(F(x))=f(x) \Rightarrow \int f(x) d x=F(x)+c$

$\frac{d}{d x}\left[\int f(x) d x\right]=\frac{d}{d x}[F(x)+c]=\frac{d F}{d x}+0$

$\frac{d}{d x} \int f(x) d x=f(x) \text {. }$

(b) $~ {f^{\prime}(x)}=\frac{d}{d x}{f(x)} \Rightarrow \int f^{\prime}(x) d x=f(x)+c , c\in \mathbb{R}$

2. $\frac{d}{d x}\left(\int f(x) d x\right)=\frac{d}{d x}\left(\int g(x) d x\right)$

$\Rightarrow \frac{d}{d x}\left[\int f(x) d x-\int g(x) d x\right]=0$

$\Rightarrow \int f(x) d x-\int g(x) d x=C$

$\{\int g(x) d x+C_1, C_1 \in \mathbb{R}\}$ & $\{\int f(x) dx+C_2, C_2 \in \mathbb{R}\}$

$\int f(x) d x=\int g(x) d x .$

$\int[f(x)+g(x)] d x=\int f(x) d x+\int g(x) d x$

Proof: $\frac{d}{d x}\left[\int(f(x)+g(x)) d x\right]$

=f(x)+g(x)......(i)

$\frac{d}{d x}\left[\int f(x) d x+\int g(x) d x\right]=\frac{d}{d x} \int f(x) d x+\frac{d}{d x} \int g(x) d x$

$=f(x)+g(x) \ldots \text { (ii) }$

$\int k f(x) d x=k \int f(x) d x \quad k \rightarrow \text { constant }$

$\frac{d}{d x} \int k f(x) d x=k f(x)$

$\frac{d}{d x}\left[k \int f(x) d x\right] =k \frac{d}{d x} \int f(x) d x =k f(x)$

$\int k f(x) d x =k \int f(x) d x$

for constants $k_1, k_3 \ldots k_n \$ functions;

$f_1(x), f_2(x), \ldots f_n(x)$,

$\int\left[k_1 f_1(x)\right. \left.+k_2 f_2(x)+\cdots+k_n f_n(x)\right] d x$

$= k_1 \int f_1(x) d x+k_2 \int f_2(x) d x+\cdots+ +k_n \int f_n(x) d x$

Example $I=\int\left(a x^2+b x+c\right) d x$

$=a \int x^2+b \int x d x+c \cdot \int 1 \cdot d x$

$I =a \frac{x^3}{3}+b \frac{x^2}{2}+c x+c_1$

$\frac{d}{d x}(\frac{x^{n+1}}{n+1})=x^n ;\quad \int x^n d x=\frac{x^{n+1}}{n+1}+c, n \neq-1$

$\frac{d}{d x}(x)=1 ;\quad\int 1 \cdot d x=x+C$

$\frac{d}{d x}(\sin x)=\cos x ; \quad\int \cos x d x=\sin x+C$

$\frac{d}{d x}(-\cos x)=\sin x ;\quad \int \sin x d x=-\cos x+C$

$\frac{d}{d x}(\tan x)=\sec ^2 x;\quad \int \sec ^2 x d x=\tan x+C$

$\frac{d}{d x}(-\cot x)=\operatorname{cosec}^2 x ;\quad \int \operatorname{cosec}^2 x d x=-\cot x+C$

$\frac{d}{d x}(\sec x)=\sec x \tan x ;\quad\int \sec x \tan x d x=\sec x+C$

$\frac{d}{d x}(-\operatorname{cosec} x)=\operatorname{cosec} x \cot x ;\quad \int \operatorname{cosec} x \cot x dx=\operatorname{-cosec} x+C$

$\frac{d}{d x}\left(\sin ^{-1} x\right)=\frac{1}{\sqrt{1-x^2}} ;\quad \int \frac{1}{\sqrt{1-x^2}} d x=\sin ^{-1} x+C$

$\frac{d}{d x}\left(-\cos ^{-1} x\right)=\frac{1}{\sqrt{1-x^2}} ;\quad \int \frac{1}{\sqrt{1-x^2}} d x=-\cos ^{-1} x+C$

$\frac{d}{d x}\left(\tan ^{-1} x\right)=\frac{1}{1+x^2} ;\quad \int \frac{1}{1+x^2} d x=\tan ^{-1} x+C$

$\frac{d}{d x}\left(-\cot ^{-1} x\right)=\frac{1}{1+x^2} ;\quad \int \frac{d x}{1+x^2}=-\cot ^{-1} x+C$

$\frac{d}{d x}\left(\sec ^{-1} x\right)=\frac{1}{x \sqrt{x^2-1}} ;\quad \int \frac{d x}{x \sqrt{x^2-1}}=\sec ^{-1} x+C$

$\frac{d}{d x}\left(-\operatorname{cosec}^{-1} x\right)=\frac{1}{x \sqrt{x-1}} ;\quad \int \frac{d x}{x \sqrt{x^2-1}}=-\operatorname{cosec}^{-1} x + C$

$\frac{d}{d x}\left(e^x\right)=e^x ;\quad \int e^x d x=e^x + C$

$\frac{d}{d x}\left(\frac{e^{n x}}{x}\right)=e^{n x} ;$

$\int e^{n x} d x=\frac{e^{n x}}{x}+C, n ≠ 0$

$\frac{d}{d x}(\log |x|)=\frac{1}{x} ;$

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

Remark: $\int e^{-x^2} d x$

$I =\int\left(4 e^{3 x}+1\right) d x$

$=4 \int e^{3 x} d x+\int 1 \cdot d x$ $\quad$ $(\int e^{3 x} d x =\frac{ e^{3 x}}{3}+c)$

$=4 \frac{e^{3 x}}{3}+4 c_1+x+c_2$

$=\frac{4}{3} e^{3 x}+x+\underbrace{4 c_1+c_2}$

I =$\frac{4}{3} e^{3 x}+x+c$

$\int \frac{x^3-x^2+x-1}{x-1} d x$

= $\int \frac{x^2(x-1)+(x-1)}{x-1} d x$

= $\int\left(x^2+1\right) d x$

= $\frac{x^3}{3}+x+C$

$I=\int f(x) d x$

x $\rightarrow$ independent Variable 't'

$x=g(t) \Rightarrow \frac{d x}{d t}=g^{\prime}(t)$

$d x=g^{\prime}(t) d t$

$I=\int f(g(t)) g^{\prime}(t) d t$

$\int f(x) d x=\int f(g(t)) g^{\prime}(t) d t$

I= $\int \frac{2 x}{1+x^2} d x$

define $t=1+x^2 \Rightarrow d t=2 x \cdot d x$.

$=\int \frac{d t}{t}=\log |t|+c$

$=\log \left|1+x^2\right|+c$

I= $\int \sin (a x+b) d x$

$a x+b=t \Rightarrow a d x=d t$

I= $\frac{1}{a} \int \sin t d t$

= $\frac{1}{a}(-\cos t)+C$

= $-\frac{\cos (a x+b)}{a}+C$