Let, I=int(x^3-2x^2+5x-3)/(-x^2+3x-4)dx,I=∫x3−2x2+5x−3−x2+3x−4dx,
=-int(x^3-2x^2+5x-3)/(x^2-3x+4)dx.=−∫x3−2x2+5x−3x2−3x+4dx.
Now, x^3-2x^2+5x-3,x3−2x2+5x−3,
=x(x^2-3x+4)+1(x^2-3x+4)+4x-7,=x(x2−3x+4)+1(x2−3x+4)+4x−7,
=(x+1)(x^2-3x+4)+(4x-7).=(x+1)(x2−3x+4)+(4x−7).
rArr (x^3-2x^2+5x-3)/(x^2-3x+4),⇒x3−2x2+5x−3x2−3x+4,
=x+1+(4x-7)/(x^2-3x+4).=x+1+4x−7x2−3x+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.!
