The integral of a function is just the opposite of the derivative of a function, hence often referred to as antiderivative, the inverse of the derivative or primitive of a function. In calculus, the integral of a function originates from the idea of finding the area under a curve by dividing it into infinitesimally small regions and then adding up or integrating all those tiny regions. Hence, integration is a type of summation when the number of terms in the sum tends to infinity, and each term of the summation is considerably small or tends to zero.

Mathematically, the integral of a function f is a function whose derivative is the function f. In other words, if F(x) is the derivative of function f, that is,

Then f is integral or primitive of F(x) such that

For example, let f = x^{2}, then derivative of f, f’(x) = 2x and ∫ 2x dx = x^{2}. But we observe that integration of 2x + 2, 2x + 5, … etc., also has the same integral, x^{2}. For this reason, whenever we find an indefinite integral, we always add a constant.

## Methods of Integration

Following are the methods of integration:

(i) Integration by substitution

(ii) Integration by partial fractions

(iii) integration by parts

### Integration by substitution

This method is used to find the integral of the type ∫ f{g(x)}.g’(x) dx. We substitute g(x) = t, then g’(x) dx = dt. The function is reduced to ∫ f(t) dt, whose integral can be easily found.

Below are the form of substitutions:

(i) When the integrand is of type ∫ f(ax + b) dx, put ax + b = t, then dx = (1/a)dt.

(ii) When the integrand is of type ∫ [f(x)]^{n}.f’(x) dx, put f(x) = t, then f’(x)dx = dt.

(iii) When the integrand is of the type ∫ x^{n – 1}.f(x^{n}) dx, put x^{n} = t, then x^{n – 1}dx = (1/n)dt.

(iv) When the integrand is of the type ∫ f’(x)/f(x) dx, put f(x) = t, then f’(x)dx = dt.

For example, we need to find the integral ∫ sec^{2} (6x + 5)dx

Let 6x + 5 = t ⇒ 6dx = dt ⇒ dx = (1/6)dt, the function becomes

⅙ ∫ sec^{2} t dt = ⅙ tan t + C

∴ ∫ sec^{2} (6x + 5) dx = ⅙ tan (6x + 5) + C

### Integration by partial fractions

This method is used when the integrand can be decomposed into partial fractions. Then each partial fraction can be integrated using standard integrals.

Learn more about integration by partial fractions.

Below are the partial fraction forms we often use while integrating partial fractions.

Form of Rational Function | Partial Fraction Decomposition |

Where x^{2} + bx + c cannot be factorised |

For example, we have to find the integral of (x + 1)/(x^{2} – 5x + 6).

First, we must factorise the denominator, x^{2} – 5x + 6; we can factorise it by splitting the middle term.

The prime factors of 6 = 2 × 3 such that 3 + 2 = 5

∴ x^{2} – 5x + 6 = (x – 2)(x – 3)

Let,

Comparing the coefficients of x and the constant term of the numerator on both sides, we get,

A = –3 and B = 4

Thus,

Integrating we get,

Hence, the use of partial fractions made the integration easy.

### Integration by parts

Integration by parts is used to integrate the product of functions, using the rule:

f(x) is the first function and g(x) is the second function. Generally, the function which vanishes by differentiating several times is taken as the first function, like x^{n}. In case logarithmic or trigonometric functions are there in the integrand, we take it as the first function, and if the second function is not there, we take it as 1.

For example, ∫ x sec^{2} x dx = [x ∫ sec^{2} x dx] – ∫[d/dx (x). ∫ sec^{2} x dx] dx + C

⇒ ∫ x sec^{2} x dx = x tan x – ∫[1. tan x] dx + C

⇒ ∫ x sec^{2} x dx = x tan x – log_{e} |cos x| + C.