INTEGRATION DECOMPOSITION METHOD

Integrate the functions with respect to x :

Example 1 :

(x³ + 4x² - 3x + 2)/x²

Solution :

= ∫[(x³ + 4x² - 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))² dx

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

= ∫[(√x)² + (1/√x)² + 2√x(1/√x)]dx

= ∫xdx + ∫(1/x)dx + 2∫dx

= (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 - 180∫dx

= (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

Example 6 :

∫4 sec² x dx

Solution :

∫4 sec² x dx = 4∫sec² x dx

= 4 tan x + c

Example 7 :

∫e^x / (1 + e^x) dx

Solution :

= ∫e^x / (1 + e^x) dx

Let u = 1 + e^x

u - 1 = e^x

du = e^x dx

∫e^x / (1 + e^x) dx = ∫(1/u) du

= log u + C

= log (1 + e^x) + C

Example 8 :

∫sin (ln x) dx

Solution :

= ∫sin (ln x) dx

Let u = sin (ln x) and dv = dx

du = -cos (ln x)(1/x) and v = x

 ∫u dv = uv - ∫v du

= sin (ln x) x + ∫x cos (ln x)(1/x) dx

= x sin (ln x) + ∫cos (ln x) dx -----(1)

Let u = cos (ln x) and dv = dx

du = sin (ln x)(1/x) and v = x

 ∫u dv = uv - ∫v du

= x cos (ln x) -  ∫x sin (ln x)(1/x) dx

∫cos (ln x) dx = x cos (ln x) -  ∫sin (ln x) dx

Applying these values in (1), we get

∫sin (ln x) dx = x sin (ln x) + x cos (ln x) -  ∫sin (ln x) dx

∫sin (ln x) dx + ∫sin (ln x) dx = x sin (ln x) + x cos (ln x)

2∫sin (ln x) dx = x sin (ln x) + x cos (ln x)

∫sin (ln x) dx = (1/2) [x sin (ln x) + x cos (ln x)]

Example 9 :

∫x²/(x³ + 1) dx

Solution :

= ∫x²/(x³ + 1) dx

Let u = x³ + 1

du = 3x² dx

x² dx = du/3

 ∫x²/(x³ + 1) dx = ∫(1/u) (du/3)

= (1/3) ∫(1/u) du

= (1/3) log u + C

= (1/3) log (x³ + 1) + C

Example 10 :

∫cos x /(1 + sin x) dx

Solution :

= ∫cos x /(1 + sin x) dx

Let u = 1 + sin x

du = - cos x dx

cos x dx = -du

= ∫-du /u

= -log u + C

= -log (1 + sin x) + C

= log (1 + sin x)^-1 + C

= log [1/(1 + sin x)] + C

Example 10 :

∫x²  e^{x^3} dx

Solution :

= ∫x²  e^{x^3} dx

Let u = x³ 

du = 3 x²  dx

x²  dx =  du/3

∫x²  e^{x^3} dx = ∫ e^u du/3

= (1/3) e^u + C

= (1/3) e^{x^3}  + C

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