SOLVED EXAMPLES OF INTEGRATION

Question 1 :

Integrate the following functions with respect to x :

e^{x log a} e^x

Solution :

∫[e^{x log a} e^x] dx

  =  ∫e^{log a^x} e^x dx

  =  ∫x e^x dx

u = x, dv  =  e^x

du  =  dx, v = e^x

∫ udv  =  uv - ∫v du

 ∫x e^x dx  =  x(e^x) - ∫ e^x dx

  =  xe^x - e^x + c

  =  x (e^x -1) + c

Question 2 :

Integrate the following functions with respect to x :

(3x + 4)√(3x + 7)

Solution :

∫(3x + 4)√(3x + 7) dx

Let t = 3x + 7

3x  =  t - 7

3x + 4  =  t - 7 + 4  ==> t - 3

Differentiating with respect to "x", we get

dt  =  3 dx  ==> dx  =  dt/3

∫(3x + 4)√(3x + 7) dx  =  ∫(t - 3)√t (dt/3)

  =  (1/3)[∫ t√t dt - 3∫√t dt]

  =  (1/3)[∫ t^3/2 dt - 3∫t^1/2 dt]

  =  (1/3)[(2/5) t^5/2 - 3 (2/3)t^3/2]

  =  (1/3)[(2/5) (3x + 7)^5/2 - 2 (3x + 7)^3/2] + c

Question 3 :

Integrate the following functions with respect to x :

(8^{1 + x} + 4^{1- x}) / 2^x

Solution :

∫(8^{1 + x} + 4^{1- x}) / 2^x dx

  =  ∫(2^3(1+x) + 2^2(1-x) / 2^x dx

  =  ∫(2^3+3x + 2^2-2x) / 2^x dx

  =  ∫(2^3+3x-x  + 2^2-2x-x) dx

  =  ∫(2^3+2x  + 2^2-3x) dx

  =  (2^3+2x/2log 2)  - (2^2-3x/3 log 2) + c

  =  (2^2+2x/log 2)  - (2^2-3x/3 log 2) + c

Question 4 :

Integrate the following functions with respect to x :

1/(√(x + 3) - √(x + 4))

Solution :

∫ 1/(√(x + 3) - √(x + 4)) dx

  =  ∫(1/(√(x + 3) - √(x + 4)) dx

Rationalizing the denominator, we get

  =  ∫[(√(x + 3) + √(x + 4))/(x + 3) - (x + 4))]dx

  =  -[∫(√(x + 3) + √(x + 4)) dx 

  =  - [∫(√(x + 3) dx + ∫√(x + 4)) dx]

  =  (2/3)(x + 3)^3/2 + (2/3)(x + 4)^3/2 + c

Question 5 :

Integrate the following functions with respect to x :

(x + 1)/(x + 2)(x + 3)

Solution :

∫ [(x + 1)/(x + 2)(x + 3)] dx

We have to use the concept partial fraction in order to integrate.

(x + 1)/(x + 2)(x + 3)  =  A/(x + 2) + B/(x + 3)

x + 1  =  A(x + 3) + B(x + 2)

Plug x = -2

-1  =  A  ===> A  =  -1

Plug x = -3

-2  =  -B  ===> B  =  2

∫(x+1)/(x+2)(x+3) dx =  -∫1/(x + 2) dx + ∫ 2/(x + 3) dx

  = - log (x + 2) + 2 log (x + 3) + c

  =  2 log (x + 3) - log (x + 2) + c

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