# INTEGRATION DECOMPOSITION METHOD

Sometimes it is difficult to integrate a function directly. But it can be integrated after decomposing it into a sum or difference of number of functions whose integrals are already known.

In most of the cases the given integrand will be any one of the algebraic, trigonometric or exponential forms, and sometimes combinations of these functions.

In the examples given below, integrate the functions with respect to x :

Example 1 :

(x3 + 4x- 3x + 2)/x2

Solution :

= ∫[(x3 + 4x- 3x + 2)/x2]dx

∫(x3/x2)dx + 4∫(x2/x2)dx - 3∫(x/x2)dx + 2∫(1/x2)dx

∫xdx + 4∫dx - 3∫(1/x)dx + 2∫x-2dx

x2/2 + 4x - 3logx - 2x-1

x2/2 + 4x - 3logx - (2/x) + c

Example 2 :

(√x + (1/√x))2

Solution :

= ∫(√x + (1/√x))dx

Expanding this using the formula (a + b)2  =  a2 + 2ab + b2

= ∫[(√x)2 + (1/√x)+ 2√x(1/√x)]dx

= ∫xdx + (1/x)dx + 2dx

= (x2/2) + logx + 2x + c

Example 3 :

(2x - 5)(36 + 4x)

Solution :

= ∫(2x - 5)(36 + 4x)dx

∫(72x + 8x2 - 180 - 20x)dx

∫(8x2 + 52x - 180)dx

∫8x2dx + 52xdx - 180dx

(8/3)x3 + 26x2 - 180x + c

Example 4 :

(cot2x + tan2x)

Solution :

= ∫(cot2x + tan2x) dx

(cosec2x - 1 + sec2x - 1)dx

cosec2xdx sec2xdx - 2dx

= -cotx + tanx - 2x + c

Example 5 :

(cos2x - cos2a)/(cosx - cosa)

Solution :

= ∫[(cos2x - cos2a)/(cosx - cosa)]dx

= ∫[(2cos2x - 1) - (2cos2a - 1)/(cosx - cosa)]dx

∫[2cos2x - 1 - 2cos2a + 1)/(cos x - cos a)]dx

∫[2(cos2x - cos2a)/(cosx - cosa)]dx

∫[2(cosx - cosa)(cosx + cosa)/(cosx - cosa)]dx

= 2(cosx + cosa)dx

= 2(sinx + xcosa) + c

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