SUBJECT: INTEGRAL CALCULUS FOR ENGINEERS

TOPIC: Integrals involving Quadratic Expressions

NOTE: Follow the formula and I need full complete solution and correct. Thanks you so much. Please help me!!

ACTIVITY # 5 (I NEED FULL COMPLETE SOLUTION AND READABLE.

GENERAL FORMULA WITH EXAMPLE (GUIDE FOR ACTIVITY):

Lesson 13 Integrals involving Quadratic Expressions When the integrand contains a quadratic expression ax2 + bx + c, where a, b and c are constants, one approach is to transform the quadratic expression into a sum or difference of a perfect square trinomial with a constant. ( ax i- b)2 i c2 The following examples will show how to transform a quadratic expression into a sum or difference of a perfect trinomial and a constant. Lesson 13 - Integrals involving Quadratic Expressions Example 1: [ 1 dx / ( x2 + 3x + 5 ) = [ dx / (x2 + 3x + ? ) +(5- ?) -> add and subtract a value in ? = [ dx / [x2 + 3x + (3/2) 2 ] + (5-(3/2)2) ->we divide 2nd term by 2 then square it = [ dx / (x2 + 3x + 9/4) + (5-9/4) transformation = [ dx / (x + 3/2)2 + 11/4 Let u = x + 3/2 du = dx a2 = 11/4 a =v11 / 2 = [ du / (u2 + a2) 1 14 = (1/a) arctan (u/a) + C = ( 1/ V11/2 ) arctan [ (x + 3/2) / v11/2 ] + C = ( 2/V11) arctan [ 2 (x + 3/2) /v11 ] + C = ( 2/V11) arctan [ (2x + 3) /V11 ] + cLesson 13 - Integrals involving Quadratic Expressions Example 2: [ 1 dx / V ( 1 + 4x - 1x2 ) 113 = [ dx / V 1( x2 4x - 1 ) +1x2 +/- bx +/- c = [ dx / V-1 [ (x2 4x + ? ) + ( 1 -? ) ] = [ dx / V -1[ ( x2 - 4x + 4 ) + (-1-4) ] = [ dx / V 1[ (x-2)2 + (-5 )] = [ dx / V [ 5 -(x-2)2 ] Let u = x - 2 du = dx a 2 = 5 a =V5 = [ du / V ( a2 - uz) 1 13 = arcsin ( u/a ) + C = arcsin [ ( x - 2 )/V5 ] + C final ans