# Complete 100+ Integration Formulas Basic to Advanced

Integration Formulas are provided here . These formulas will help the student to quickly revise their subject of Integration. These are helpful in exams

Here are the complete 100+ Integration formulas. This post contains all the integration formulas from basic to Advance.

Contents

## Integrals / Anti Derivative

A function ϕ(x) is said to be anti derivative of a function if ϕ'(x) = f(x).
$$\frac{d}{dx}{\varphi(x)}\ =\ f(x)$$

## Indefinite Integrals

If a f(x) is a function , then the family of all its antiderivatives is called Indefinite Integrals. The general form of an Indefinite Integrals is provided below :-
$$\int{f(x).dx}$$
It is called indefinite integral because it is not unique. The value of definite integrals is unique. An indefinite integral does not have any limit to apply. But a definite integral has upper limits and lower limits.
Indefinite Integrals – $$\int{f\left(x\right).dx}=\varphi\left(x\right)+C$$
Definite Integrals – $$\int_{a}^{b}{{f(x)}.dx\ }=\ {\varphi(x)}_a^b\ =\ \varphi(b)\ -\ \varphi(a)\$$

## Standard Integration Formulas

These are the basic integration formulas that are used to solve the problems of integration functions.

1. $$\int x^n.dx\ =\ \frac{x^{n+1}}{n+1}\ +\ C$$

2. $$\int1.dx\ =\ x\ +C$$

3. $$\int\frac{1}{x}.dx\ =\ \log{x}\ +C$$

4. $$\int\cos{x}\ .\ dx\ =\ \sin{x}\ +\ C$$

5. $$\int\sin{x}\ .dx\ =\ -\cos{x}\ +\ C$$

6. $$\int{\sec^2x\ }.dx\ =\ tan\ x\ +C\$$

7. $$\int{{\rm cosec}^2x}\ .\ dx\ =\ -\ cot\ x\ +\ C$$

8. $$\int{sec\ x\ tan\ x}\ .\ dx\ =\ sec\ x\ +\ C$$

9. $$\int{cosec\ x\ cot\ x\ }.dx\ =\ -\ cosec\ x\ +\ C$$

10. $$\int e^x\ .dx\ =\ e^x\ +\ C$$

11. $$\int a^x.dx\ =\ \frac{a^x}{\log{a}}\ +\ C$$

12. $$\int\frac{1}{\sqrt{1\ -\ x^2}}\ .dx\ =\ \sin^{-1}{\frac{x}{1}}\ +C$$

13. $$\int\frac{1}{\sqrt{a^2\ -\ x^2}}\ .dx\ =\ \sin^{-1}{\frac{x}{a}}\ +C$$

14. $$\int{\frac{1}{1+x^2}\ }.dx\ =\ \tan^{-1}{x}\ +\ C$$

15. $$\int{\frac{1}{a^2\ +\ x^2}\ }.dx\ =\ \frac{1}{a}\ \tan^{-1}{\frac{x}{a}}\ +\ C$$

16. $$\int{\frac{1}{x\sqrt{x^2-1}}\ .dx}=\ \sec^{-1}{x}\ +\ C$$

17. $$\int{-\frac{1}{x\sqrt{x^{2\ }-\ 1}}.dx=}{\rm cosec}^{-1}{x}+\ C$$

18. $$\int{-\frac{1}{a^2\ +\ x^2}.dx\ =\ \frac{1}{a}\ }\cot^{-1}{\frac{x}{a}}\ +\ C$$

19. $$\int{\frac{1}{x\sqrt{x^2\ -\ a^2}}\ .dx}\ =\ \frac{1}{a}\ \sec^{-1}{\frac{x}{a}}\ +\ C$$

20. $$\int{-\frac{1}{x\sqrt{x^2\ -\ a^2}}\ .dx}\ =\ \frac{1}{a}\ co\sec^{-1}{\frac{x}{a}}\ +\ C$$

## Integration by Substitution

The method of evaluating integrals of a function by substituting a suitable function is known as integration by substitution . Here are some of the integrals which are used in the method of Integration by substitution. In these fundamental integrals, x is replace by ” ax + b ” .

### Integration Formulas For Substitution method

1. $$\int{\left(ax+b\right)^n\ .dx\ }=\ \frac{\left(ax+b\right)^{n+1}}{a\left(n+1\right)}\ +C\$$

2. $$\int{\frac{1}{ax\ +\ b\ }\ .dx}\ =\frac{1}{a}\ \log{\left(ax+b\right)}\ +C\$$

3. $$\int{e^{\left(ax+b\right)}.dx}\ =\ \frac{1}{a}e^{\left(ax+b\right)}\ +\ C$$

4. $$\int{a^{\left(ax+b\right)}.dx}\ =\ \frac{1}{b}\times\frac{a^{\left(bx+c\right)}}{\log{a}}\ +\ C$$

5. $$\int\sin{\left(ax+b\right).dx}$$
$$=-\frac{1}{a}cos\left(ax+b\right)+C$$

6. $$\int{cos\left(ax+b\right).dx}$$
$$=\frac{1}{a}sin\left(ax+b\right)+C$$

7. $$\int{\sec^2\left(ax+b\right).dx}$$
$$=\frac{1}{b}tan\left(ax+b\right)+C$$

8. $$\int{{\rm cosec}^2\left(ax+b\right).dx}$$
$$=-\frac{1}{a}cot\left(ax+b\right)+C$$

9. $$\int{sec\left(ax+b\right)tan\left(ax+b\right).dx}$$
$$=\frac{1}{a}sec\left(ax+b\right)+C$$

10. $$\int{cosec\left(ax+b\right)cot\left(ax+b\right).dx}$$
$$=-\frac{1}{a}cosec\left(ax+b\right)+C$$

11. $$\int{tan\left(ax+b\right).dx}\$$
$$=\ -\frac{1}{a}log\left|cos\left(ax+b\right)\right|\ +C\$$

12. $$\int{cot\left(ax+b\right).dx}$$
$$=\frac{1}{a}log\left|sin\left(ax+b\right)\right|\ +\ C\$$

13. $$\int{tan\ x.dx}\$$
$$=\ -\ log\left|cos\ x\right|\ +\ C\ \$$
$$=\ log\left|sec\ x\right|\ +\ C$$

14. $$\int{cot\ x\ .\ dx}\ =\ log\left|sin\ x\right|\ +\ C$$

15. $$\int{sec\ x\ .dx\ }\$$
$$=\ log\left|sec\ x\ +\ tan\ x\right|\ +\ C\$$

16. $$\int{sec\ x\ .dx\ }\$$
$$=log\left|\tan{\left(\frac{\pi}{2}\ +\ \frac{x}{2}\right)}\right|\ +\ C\$$

17. $$\int{cosec\ x\ .dx}\$$
$$=\ log\left|cosec\ x\ -\ cot\ x\right|\ +\ C\$$
$$=\ log\left|tan\ \frac{x}{2}\right|\ +\ C$$

### Steps to Solve Integration Questions by Substitution Method

• Select a new variable , say y =g(x)
• Now after selecting the new variable , find out the value of dx , by using the $$\frac{dy}{dx}$$ = g'(x) .This implies
dy = g'(x)dx
• Now put the value of given integral in the form of another variable.
• Also change the value of all variable w.r.t the new assumed variable.
• Integrate the new formed integral.
• Replace the value of assumed variable with the original one.

Read Also : Trigonometry Formulas

## Integration by Partial Fractions

Any rational number of the the form $$\frac{P\left(x\right)}{Q\left(x\right)}$$ can by solved by the converting them into some standard partial fraction form . In this method we have to convert the $$\frac{P\left(x\right)}{Q\left(x\right)}$$ to a partial fraction form. Some standard Result are shown in the table below.

## Integration by Parts

When we have to functions which are in multiplication with each other and their integral value is to be find , then we use integration by parts method. The integration by parts method is as follows

$$\int{u.v}\ dx\ =\ u.\int v.dx\ -\int{{\ \frac{du}{dx}\ .\ \int v.dx}}\ .\ dx$$

### The ILATE Concept

To integrate the product of the two functions ( u and v ) , we have decide which function is to be taken as u and which one as v , This Can be easily done by ILATE Concepts.
The ILATE stands for –
I – Inverse Trigonometric functions
L – Logarithmic functions
A – Algebraic functions
T – Trigonometric Functions
E – Exponential Functions .

If we have two functions then we have to chose the function ‘u’ which is earlier in the ILATE concept and the other as ‘v’ .

## Some Standard Integral Values

1. $$\int{\frac{1}{x^2\ +\ a^2}\ .dx\ }=\ \frac{1}{a}\ \tan^{-1}{\frac{x}{a}}\ +\ C$$

2. $$\int{\frac{1}{x^2\ -\ a^2}\ .dx}\ =\ \frac{1}{2a}\ \log{\left|\frac{x\ -\ a\ }{x\ +\ a}\right|}\ +\ C$$

3. $$\int{\frac{1}{a^2\ -\ x^2}\ .dx\ }\ =\ \frac{1}{2a}\ \log{\left|\frac{a+x}{a-x}\right|}\ +\ C$$

4. $$\int{\frac{1}{\sqrt{x^2\ +\ a^2}}\ .\ dx\ }\ =\ \log{\left|\frac{x\ +\ \sqrt{x^2\ \ +\ a^2}}{a}\right|}\ +\ C\$$

5. $$\int{\frac{1}{\sqrt{x^2\ -\ a^2}}\ .\ dx\ }\ =\ \log{\left|\frac{x\ +\ \sqrt{x^2\ \ -\ a^2}}{a}\right|}\ +\ C\$$

6. $$\int{\frac{1}{\sqrt{a^2\ -\ x^2}}\ .\ dx\ }\ =\ \sin^{-1}{\frac{x}{a}\ }\ +\ C\$$

7. $$\int{\sqrt{a^2\ -\ x^2}\ }.dx\ =\ \frac{x}{2}\ \sqrt{a^2\ -\ x^2}\ +\ \frac{a^2}{2}\ \sin^{-1}{\frac{x}{a}}\ +\ C$$

8. $$\int\sqrt{x^2\ -\ a^2}\ .dx=\ \frac{x}{2}\ \sqrt{x^2\ -\ a^2}\ -\ \frac{a^2}{2}\ \log{\left|x\ +\ \sqrt{x^2\ -\ a^2\ }\right|}+\ C\ \$$

9. $$\int\sqrt{x^2\ +\ a^2}\ .dx=\ \frac{x}{2}\ \sqrt{x^2\ +\ a^2}\ +\ \frac{a^2}{2}\ \log{\left|x\ +\ \sqrt{x^2\ +\ a^2\ }\right|}+\ C$$