Question

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

Answer 1

`-(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.!