How do you integrate #\int \frac { x ^ { 2} - 1} { 4x ^ { 2} + 6x + 9}#?

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Answer 1 · 2018-03-03T18:56:47

# 1/4x-3/16ln(4x^2+6x+9)-17/(24sqrt3)arc tan{(4x+3)/(3sqrt3)}+C.#

Let, #I=int(x^2-1)/(4x^2+6x+9)dx#,

#=1/4int(4x^2-4)/(4x^2+6x+9)dx#,

#=1/4int{(4x^2+6x+9)-(6x+13)}/(4x^2+6x+9)dx#,

#=1/4int{(4x^2+6x+9)/(4x^2+6x+9)-(6x+13)/(4x^2+6x+9)}dx#,

#=1/4int{1-(6x+13)/(4x^2+6x+9)}dx#,

#=1/4x-1/4int(6x+13)/(4x^2+6x+9)dx#,

#=1/4x-6/4int(x+13/6)/(4x^2+6x+9)dx#,

#=1/4x-6/4*1/8int(8x+8*13/6)/(4x^2+6x+9)dx#,

#=1/4x-3/16int(8x+52/3)/(4x^2+6x+9)dx#,

#=1/4x-3/16int{(8x+6)+34/3}/(4x^2+6x+9)dx#,

#=1/4x-3/16int{(8x+6)/(4x^2+6x+9)+34/3*1/(4x^2+6x+9)}dx#,

#=1/4x-3/16int{d/dx(4x^2+6x+9)}/(4x^2+6x+9)dx#

#-3/16*34/3int1/(4x^2+6x+9)dx#,

#=1/4x-3/16ln(4x^2+6x+9)#

#-17/8int1/{(4x^2+6x+9/4)+27/4}dx#,

#=1/4x-3/16ln(4x^2+6x+9)#

#-17/8int1/{(2x+3/2)^2+((3sqrt3)/2)^2}dx#,

#=1/4x-3/16ln(4x^2+6x+9)#

#-17/8*2/(3sqrt3)*1/2arc tan((2x+3/2)/(3sqrt3/2))#.

#rArr I=1/4x-3/16ln(4x^2+6x+9)-17/(24sqrt3)arc tan{(4x+3)/(3sqrt3)}+C.#