ONE MORE STEP TOWARD BETTER TOMORROW : CHAPTER 8- Worked Out Examples

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)

	$4{x^2} + 5x + 4 = 4\left( {{x^2} + \dfrac{5}{4}x + 1} \right)$
	$= 4\left( {{{\left( {x + \dfrac{5}{8}} \right)}^2} + 1 - {{\left( {\dfrac{5}{8}} \right)}^2}} \right)$
	$= 4{\left( {x + \dfrac{5}{8}} \right)^2} + \dfrac{{39}}{{16}}$
	$= {\left( {2x + \dfrac{5}{4}} \right)^2} + {\left( {\dfrac{{\sqrt {39} }}{4}} \right)^2}$