Evaluate the integral. (Remember to use absolute values where appropriate. Use C for the constant of integration.)

Question

	5x3 + 20x2 + 32x − 1
(x2 + 4x + 5)2
dx

Answer 1

Firstly decompose the fraction ( 5x3 + 20x2 + 32x − 1 ) / (x2 + 4x + 5)2 . Dy doing it you will get

( 5x3 + 20x2 + 32x − 1 ) / (x2 + 4x + 5)2 = 5x / ( x2 + 4x + 5) + ( 7x -1 ) / (x2 + 4x + 5)2.

A. 5x / ( x2 + 4x + 5) = 5 [( x +2- 2) / ( x2 + 4x + 5) = 5( x+2 )/ ( x2 + 4x + 5) - 10 /( x2 + 4x + 5) =

(5/2) ( 2 x + 4 ) / ( x2 + 4x + 5) - 10 / [ (x+2)2 +1]

Therefore ∫[5x / ( x2 + 4x + 5)] d x = ∫ [ (5/2) ( 2 x + 4 ) / ( x2 + 4x + 5) - 10 / [ (x+2)2 +1] ] d x=

∫ [ (5/2) ( 2 x + 4 ) / ( x2 + 4x + 5)]d x -10 ∫ d x / [ (x+2)2 +1] =

(5/2) ln (x2 + 4x + 5) - 10 tan-1 ( x+2 ).

B. ( 7x -1 ) / (x2 + 4x + 5)2.= [7(x +2) - 15]/ (x2 + 4x + 5)2.= 7(x +2) / (x2 + 4x + 5)2 - 15/ (x2 + 4x + 5)2

= ( 7/2)(2x +4) / (x2 + 4x + 5)2 - 15/[ (x + 2)2+ 1]2 Hence

∫ [ ( 7/2)(2x +4) / (x2 + 4x + 5)2 - 15/[ (x + 2)2+ 1]2 ] d x =

∫ ( 7/2)(2x +4) d x / (x2 + 4x + 5)2 -15 ∫ d x/ [ (x + 2)2+ 1]2 ] =

-( 7/2) / (x2 + 4x + 5) - 15 ∫ d u / ( u2 +1 )2 ( u = x+2 )

-( 7/2) / (x2 + 4x + 5) - 15 ∫ d u / ( u2 +1 )2 ( u = tan θ )

-( 7/2) / (x2 + 4x + 5) - 15 ∫ sec2θ d θ / sec4θ =

-( 7/2) / (x2 + 4x + 5) - 15 ∫ cos2θ d θ =

-( 7/2) / (x2 + 4x + 5) - (15/2) ∫2 cos2θ d θ =

-( 7/2) / (x2 + 4x + 5) - (15/2) ∫[cos2θ +1] d θ =

-( 7/2) / (x2 + 4x + 5) - (15/2) ∫[cos2θ ] dθ -∫15/2∫] d θ =

-( 7/2) / (x2 + 4x + 5) - (15/4) [sin2θ ] -15/2 θ =

-( 7/2) / (x2 + 4x + 5) - (15/2) [sinθcos θ ] -15/2 θ =

-( 7/2) / (x2 + 4x + 5) - (15/2) [u/ (u2 +1)]-15/2tan -1 (x +2 )

-( 7/2) / (x2 + 4x + 5) - (15/2) [(x +2) / (x2 +4x +5)]-15/2tan -1 (x +2 )

And finally

∫ [( 5x3 + 20x2 + 32x − 1 ) / (x2 + 4x + 5)2] dx =

= (5/2) ln (x2 + 4x + 5) - 10 tan-1 ( x+2 ).-( 7/2) / (x2 + 4x + 5) - (15/2) [(x +2) / (x2 +4x +5)]-15/2tan -1 (x +2 )

= (5/2) ln (x2 + 4x + 5) .-( 7/2) / (x2 + 4x + 5) - (15/2) [(x +2) / (x2 +4x +5)] -35/2tan -1 (x +2 )

= (5/2) ln (x2 + 4x + 5) .+(15+37x)/ {2(x2 + 4x + 5) } -35/2tan -1 (x +2 )