# Indefinite Integral

If 'f' and 'g' are functions of 'x', such that g'(x)=f(x) then the function 'g' is called an integral of 'f' with respect to 'x', and is written symbolically as: f(x)dx = g(x) + c

where: f(x) is called the integrand and 'c' is called the constant of integration

Note: If d/dx f(x) = g(x) then d/dx {f(x) + c} = g(x)

Where 'c' is constant, because differentiation of a constant is zero.

Thus the general value g(x)dx is f(x)+c, where 'c' is the constant of integration.

Clearly integral will change if 'c' changes. Thus integral of a function is not unique, hence it is called indefinite integral.

### Standard Results:

These standard results for integral calculus are derived directly from the standard results of differential calculus

Differential Calculus Integral Calculus
d/dx(xn+1/ n+1) = xn xn dx =(xn+1/n+1) + C [n not =1]
d/dx loge|x| = 1/x 1/x dx = loge|x| + c [n= -1]
d/dx ex = ex ex dx = ex + c
d/dx ax = ax logea ax dx = ax / logea + c [a>0]
d/dx Cosx = - Sinx Sinx dx = - Cosx +c
d/dx Sinx = Cosx Cosx dx = Sinx + c
d/dx Tanx = Sec2x Sec2x dx = Tanx + c
d/dx Cotx = - Cosec2 x Cosec2x dx = - Cotx + C
d/dx Secx = Secx.Tanx Secx.Tanx dx = Secx + c
d/dx Cosecx = - Cosecx.Cotx Cosec.Cotx dx = - Cosecx + c
d/dx Sin-1x = 1/v(1-x2) 1/(1-x2) dx = Sin-1 + c
d/dx Tan-1x = 1/(1+x2) 1/(1+x2) dx = Tan-1x + c
d/dx Sec-1x = 1/xv(x2 - 1) 1/(x2 - 1) dx = Sec-1 x + C
d/dx Sin-1x/a = 1/v(a2 + x2) dx/v(a2 - x2) = Sin-1x/a + c
d/dx (1/a) Tan-1x/a = 1/(x2+a2) dx/(x2+a2) = 1/a Tan-1(x/a) +c
d/dx (1/a Sec-1x/a) =1/xv(x2 - a2) dx/xv(x2-a2) = 1/a Sec-1x/a +c
d/dx Coshx = Sinhx Sinh dx = Coshx + c
d/dx Sinhx = Coshx Coshx dx = Sinhx + c
d/dx Tanhx = Sech2x Sec2x dx = Tanhx + c
d/dx Cothx = - Cosech2x Cosech2x dx = - Cothx +c
d/dx Sechx = - Sechx.Tanhx Sechx.Tanhx dx = - Sechx + c
d/dx Cosechx = - Cosechx.Cothx Cosechx.Cothx dx= -Cosechx+c