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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 $x$