\(\int f(x)\;dx=F(x)\) means that \(F'(x)=f(x)\). We say that \( 'F'\)  is the indefinite integral of  \( 'f'\).

• \(\int f(x)\;dx\) is the anti derivative function.

• Indefinite integral represent an entire family of curves.

\(\int x^3\;dx=\dfrac {x^4}{4}+C\rightarrow\) each value of 'C' will give an anti-derivative.

\(\therefore \)  \(\dfrac {x^4}{4}+1,\;\dfrac {x^4}{4}-2,\;\;;\dfrac {x^4}{4}-6\) are all anti-derivatives of \(x^3\).

\(\int \limits_a^b f(x)dx=\int f(x)dx\Big]_a^b\)

• \(\int_a^b\;f(x)\;dx\) is a value whereas \(\int f(x)dx\) is a function.

Consider 

\(I=\int\;x^2\;dx=\underbrace {\dfrac {x^3}{3}+C}_{function \;of \;x}\)

whereas \(\int\limits_1^2\;x^2\;dx=\dfrac {x^3}{3} \Bigg]_1^2=\dfrac {8}{3}-\dfrac {1}{3}=\underbrace{\dfrac {7}{3}}_{value}\)

(1) \(\int c\,f(x)\;dx=c\int\,f(x)\;dx\)

(2) \(\int \Big(f(x)+g(x)\Big)\;dx=\int f(x)dx+\int g(x)\;dx\)

(3) \(\int x^n\;dx=\dfrac {x^n+1}{n+1}+C\) (for all constants \(n\))

(4) \(\int k\;dx=kx+C\) where \('k'\) is a constant

\(\displaystyle\int_a^b\,f(x)\;dx=F(b)-F(a)\)  where \(\displaystyle\int f(x)=F(x)+C\)

The integration constant 'C' has no role in the definite integral as it cancels out.

\(\displaystyle\int_a^b\,f(x)\;dx=F(x)+C\Bigg]_a^b\)

\(=\Big(F(b)+C\Big)-\Big(F(a)+C\Big)\)

\(=F(b)+C-F(a)-C\)

\(=F(b)-F(a)\)

Rules of Indefinite Integrals:

(1) \(\int c\,f(x)\;dx=c\int\,f(x)\;dx\)

(2) \(\int \Big(f(x)+g(x)\Big)\;dx=\int f(x)dx+\int g(x)\;dx\)

(3) \(\int x^n\;dx=\dfrac {x^n+1}{n+1}+C\) (for all constants \(n\))

(4) \(\int k\;dx=kx+C\) where \('k'\) is a constant

Indefinite Integrals of Trigonometric Functions

(1) \(\displaystyle\int\,sin\,x\;dx=-cos\,x+C\)

(2) \(\displaystyle\int\,cos\,x\;dx=sin\,x+C\)

(3) \(\displaystyle\int\,sec^2\,x\;dx=tan\,x+C\)

(4) \(\displaystyle\int\,cosec^2\,x\;dx=-cot\,x+C\)

(5) \(\displaystyle\int\,sec\,x\;tan\,x\;dx=sec\,x+C\)

(6) \(\displaystyle\int\,cosec\,x\;cot\,x\;dx=-cosec\,x+C\)

All these can be verified by differentiating the right hand function.

(1) \(\displaystyle\int\,sin\,x\;dx=-cos\,x+C\)

(2) \(\displaystyle\int\,cos\,x\;dx=sin\,x+C\)

(3) \(\displaystyle\int\,sec^2\,x\;dx=tan\,x+C\)

(4) \(\displaystyle\int\,cosec^2\,x\;dx=-cot\,x+C\)

(5) \(\displaystyle\int\,sec\,x\;tan\,x\;dx=sec\,x+C\)

(6) \(\displaystyle\int\,cosec\,x\;cot\,x\;dx=-cosec\,x+C\)

All these can be verified by differentiating the right hand function.

In some cases, it is required to split the numerator to solve the indefinite integration. 

e.g.  \(\displaystyle \int \dfrac{x^2+2x+1}{x+1}dx\) 

\(\Rightarrow \displaystyle\int \dfrac{(x+1)(x+1)}{x+1}dx\)

\(\Rightarrow \displaystyle\int (x+1)dx\)

\(\Rightarrow \dfrac{x^2}{2} +x+C\)