In calculus, an **antiderivative**, primeval or unclear integral of a function f is a task F whose imitative is equal to f, i.e., F? = f. The process of solving for antiderivatives is called antidifferentiation and its differing occupation is called differentiation, which is the development of result a derived. The antiderivative, or primeval, of a intricate-valued function g is a function whose intricate derivative is g. Antiderivatives are connected to definite integrals through the basic theorem of calculus the exact integral of a task over an interval is different to the difference between the values of an antiderivative evaluate at the endpoints of the distance.

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• The basic of a satisfactory, state for a function f to have an antiderivative is that f has the middle value property. That is, if [a, b] is a subinterval of the sphere of f and d is any actual number between f (a) and f (b), then f(c) = d for some c between a and b.

• The set of discontinuities of f should be a meager set. They can make some significance values of f having an anti-derivative, which has to locate as it’s location of discontinuity.

• If f has an antiderivative, is bound on close finite subinterval of the domain and has a set of discontinuities of Lévesque measure 0, then an antiderivative may be found by integration.

The method of finding antiderivatives is called antidifferentiation or integration:

d/dx [F(x)] = f(x)

`=>` `int` f(x) dx = F(x) + C.

The expression F(x) + C is the general antiderivative of ƒ. If ƒ and g are defined on the same interval, then the general antiderivative of the sum of ƒ and g equals the sum of the general antiderivatives of ƒ and g.

`int` x^{n} dx = `(x^(n + 1))/(n + 1)` + C (for n ? - 1)

`int` `1/x` dx = ln |x| + C

`int` e^{x} dx = e^{x} + C

`int` sin x dx = - cos x + C

`int` cos x dx = sin x + C

`int` sec^{2} x dx = tan x + C

`int` cosec^{2} x dx = - cot x + C

`int` sec x tan x dx = sec x + C

`int` cosec x cot x dx = - cosec x + C

`int` tan x dx = ln |sec x| + C

`int` cot x dx = - ln |cosec x| + C

`int` sec x dx = ln |sec x + tan x| + C

`int` cosec x dx = ln |cosec x - cot x| + C

Below are the examples on **antiderivatives** -

**Example 1:** Find antiderivative for the function, f(x) = 3x^{2} + 7x.

**Solution:**

** **** Step 1: **Given function

f(x) = 3x^{2} + 7x

`int` f(x) dx = `int` (3x^{2} + 7x) dx

** Step 2: **Separate the integral function

`int`(3x^{2} + 7x) dx = `int` 3x^{2} dx + `int` 7x dx

**Step 3: **Integrate each function with respect to ' x',

`int`(3x^{2} + 7x) dx = `(3x^3)/3` + `(7x^2)/2` + C

= x^{3}+ `(7x^2)/2` + C

**Example 2:** Find antiderivative for the function, f(x) = `2/x^7`

**Solution:**

** ** **Step 1: **Given function

f(x) = `2/x^7 `

`int` f(x) dx = `int` `2/x^7` dx

** Step 2:** Integrate the function `2/x^7` with respect to ' x ',

`int` `2/x^7` dx = `int` 2x^{-7} dx

= 2`((x^(-7 + 1))/(-7 + 1))`

= 2 `((x^-6)/(-6))` + C

= - `1/(3x^6)` + C

**Example 3:** Find antiderivative of y = 2cos 8x

**Solution:**

** ** **Step 1:** Given function

y = 2cos 8x

`int` y dx = `int` 2cos 8x dx

** Step 2: **Integrate the given function y = 2cos 8x with respect to ' x',

`int`2cos 8x dx = 2sin (8x) `1/8`

= `(sin (8x))/4`