What is #int (x^3-2x^2+5x-3 ) / (-x^2+ 3 x -4 )dx#?

1 Answer
Dec 11, 2017

# -(x^2/2+x)-2ln|x^2-3x+4|+2/sqrt7*arc tan((2x-3)/sqrt7)+C.#

Explanation:

Let, #I=int(x^3-2x^2+5x-3)/(-x^2+3x-4)dx,#

#=-int(x^3-2x^2+5x-3)/(x^2-3x+4)dx.#

Now, #x^3-2x^2+5x-3,#

#=x(x^2-3x+4)+1(x^2-3x+4)+4x-7,#

#=(x+1)(x^2-3x+4)+(4x-7).#

#rArr (x^3-2x^2+5x-3)/(x^2-3x+4),#

#=x+1+(4x-7)/(x^2-3x+4).#

#:. I=-int{x+1+(4x-7)/(x^2-3x+4)}dx,#

#=-(x^2/2+x)-int(4x-7)/(x^2-3x+4)dx.......(star).#

To evaluate the Integral in the R.H.S. of #(star),# we have to set,

#4x-7=m*d/dx(x^2-3x+4)+n," where "m,n in RR; i.e.,#

we have to determine #m,n in RR,# such that,

#4x-7=m(2x-3)+n.#

Clearly, #m=2, n=-1.#

#:. int(4x-7)/(x^2-3x+4)dx,#

#=int{2*d/dx(x^2-3x+4)-1}/(x^2-3x+4)dx,#

#=2int{d/dx(x^2-3x+4)}/(x^2-3x+4)dx-int1/(x^2-3x+4)dx,#

#=2ln|(x^2-3x+4)|-int1/(x^2-3x+9/4+7/4}dx,#

#=2ln|(x^2-3x+4)|-int1/{(x-3/2)^2+(sqrt7/2)^2}dx,#

#=2ln|x^2-3x+4|-1/(sqrt7/2)*arc tan{(x-3/2)/(sqrt7/2)}.#

Altogether, we have,

#I=-(x^2/2+x)-2ln|x^2-3x+4|#
#+2/sqrt7*arc tan((2x-3)/sqrt7)+C.#

Enjoy Maths.!