Basic concepts and Standard Formulas of Indefinite Integration

  • It is a process of finding the limit of a sum of a certain number of elements, as the number of elements tends to infinity when each of the elements tends to zero.
  • Integration is a reverse process of differentiation.

If {d (F(x) + c)/dx} = f(x), then

Integration of f(x).dx = F(x) + C ; Here ‘C’ is constant of integration

F(x) is Integral of f(x) or Anti-Derivative of f(x) or Primitive of f(x)

  • In differentiation dy/dx is used as a composite symbol or as an operator. In ‘Methods of Differentiation’ these do not have separate meanings.
  • But in Integration, it is treated as a fraction where ‘dy’ is called differential of ‘y’ and ‘dx’ is called differential of ‘x’.
  • For a single variable function y=f(x), differentiation of ‘y’ is denoted as ‘dy’

i.e., dy = Product of derivative of F(x) w.r.t ‘x’ and differential of x is denoted by ‘dx’.

Ex-  y = cos x then dy = -sin x dx


y = ∫ f(x) dx denotes a family of curves such that derivative at x= x1 of every member of the family is same i.e. tangent

drawn to each member of the family at x = x1 is parallel.

Ex- y = ∫ 2x dx = x2 + C

Geometrical Interpretation
Fig.- Geometrical Interpretation


  • Anti-derivative of a continuous function is always differentiable.
  • Anti-derivative of a periodic function needn’t be a periodic function.

Integral of Standard Functions

∫ (xn)dx = (xn+1)/(n+1) + C ; n ≠ -1

∫ (ax+b)n dx = (ax+b)n+1)/a(n+1) + C ; n ≠ -1

∫1/x dx = ln |x| + C

∫ 1/(ax+b) dx = 1/a ln |ax+b| +C

∫ ax dx = ax/ln a + C ; a>0

∫ apx+q dx = apx+q/p ln a + C ; a>0

∫ ex dx = ex + C

∫ epx+q dx = (epx+q)/a  + C


∫ sin x dx = – cos x +C

∫ sin (ax+b )dx = – 1/a cos (ax+b)  + C

∫ cos x dx =  sin x +C

∫ cos (ax+b) dx = 1/a sin (ax+b)  + C

∫ sec2 x dx =  tan x +C

∫ sec2 (ax+b) dx = 1/a tan (ax+b) + C

∫ cosec2 x dx =  – cot x + C

∫ cosec2 (ax+b) dx = – 1/a cot (ax+b) + C

∫sec x tan x dx = sec x + C

∫sec (ax+b) tan (ax+b) dx = 1/a sec (ax+b) + C

∫ cosec x cot x dx = – cosec x + C

∫ cosec (ax+b) cot (ax+b) dx = – 1/a cosec (ax+b) + C


∫dx/ (1+x2) dx = tan-1 x + C

∫dx/ (a2 + x2) dx = 1/a tan-1 x + C

∫dx/ (1- x2)1/2 dx = sin-1 x + C

∫dx/ (a2 – x2)1/2 dx = 1/a sin-1 (x/a) + C

∫dx/ x( x2 – 1)1/2 dx = sec-1 x + C

∫dx/ x( x2 – a2)1/2 dx = 1/a sec-1 (x/a) + C

If there is any typo error, please do let us know.

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