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Monday, 4 August 2014

CHAPTER 8- Worked Out Examples

 Example: 1    

Evaluate the following integrals:
(a) \int {\dfrac{1}{{{x^2} + 3x + 2}}} \,\,dt
(b) \int {\dfrac{1}{{4{x^2} + 4x + 2}}} \,\,dt
(c) \int {\dfrac{1}{{\sqrt {4{x^2} + 5x + 4} }}} \,\,dt
(d) \int {\dfrac{1}{{\sqrt { - 4{x^2} - 4x} }}} \,\,dt
Solution: 1-(a)

Our attempts in each of these questions will be to reduce (rearrange) the quadratic expressions so that it resembles one of our standard forms of Table -1
{x^2} + 3x + 2 = {\left( {x + \dfrac{3}{2}} \right)^2} - \dfrac{1}{4}
 = {\left( {x + \dfrac{3}{2}} \right)^2} - {\left( {\dfrac{1}{2}} \right)^2}
Thus, \,\,\,\,\,\,\,\,\, I = \int {\dfrac{1}{{{{\left( {x + \dfrac{3}{2}} \right)}^2} - {{\left( {\dfrac{1}{2}} \right)}^2}}}} \,\,dx
This is of the form (21) where a = \dfrac{1}{2}
I = \ln \left( {\dfrac{{x + 1}}{{x + 2}}} \right) + C
Solution: 1-(b)

4{x^2} + 4x + 2 = \left( {4{x^2} + 4x + 1} \right) + 1
 = {\left( {2x + 1} \right)^2} + {1^2}
Thus, \,\,\,\,\,\,\,\, I = \int {\dfrac{1}{{{{\left( {2x + 1} \right)}^2} + {1^2}}}} \,\,dx
This is of the form (17) which contains the expression {X^2} + {A^2}X = 2x + 1\,\, and A = 1
Thus,
I = \overbrace{\dfrac{1}{2}}^{Comment}{\tan ^{ - 1}}(2x + 1) + C
Comment: We have to divide by the coefficient of x
Solution: 1-(c)

4{x^2} + 5x + 4 = 4\left( {{x^2} + \dfrac{5}{4}x + 1} \right)
 = 4\left( {{{\left( {x + \dfrac{5}{8}} \right)}^2} + 1 - {{\left( {\dfrac{5}{8}} \right)}^2}} \right)
 = 4{\left( {x + \dfrac{5}{8}} \right)^2} + \dfrac{{39}}{{16}}
 = {\left( {2x + \dfrac{5}{4}} \right)^2} + {\left( {\dfrac{{\sqrt {39} }}{4}} \right)^2}
The integral is therefore of the form (22). The result is
I = \overbrace{\dfrac{1}{2}}^{Comment}\ln \left| {\left( {2x + \dfrac{5}{4}} \right) + \sqrt {4{x^2} + 5x + 4} } \right| + C
Comment: again division by the coefficient of x
Solution: 1-(d)

 - 4{x^2} - 4x = 1 - \left( {4{x^2} + 4x + 1} \right)
 = {1^2} - {\left( {2x + 1} \right)^2}
The integral is therefore of the form (15)
The result is
I = \overbrace{\dfrac{1}{2}}^{Comment}{\sin ^{ - 1}}(2x + 1) + C
Comment: again division by the coefficient of x
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