How do you integrate ((3x^2)-25x+43)dx/((2x+1)((x-2)^2))?
1 Answer
Explanation:
Lets rewrite integrand
(3x^2-25x+43)/((2x+1)(x-2)^2)=A/(2x+1)+B/(x-2)+C/(x-2)^2=
=(A(x-2)^2+B(2x+1)(x-2)+C(2x+1))/((2x+1)(x-2)^2)=
=(A(x^2-4x+4)+B(2x^2-3x-2)+C(2x+1))/((2x+1)(x-2)^2)=
=(Ax^2-4Ax+4A+2Bx^2-3Bx-2B+2Cx+C)/((2x+1)(x-2)^2)=
=(x^2(A+2B)+x(-4A-3B+2C)+(4A-2B+C))/((2x+1)(x-2)^2)
Comparing with integrand coefficients we get system of linear equations with 3 unknowns:
A+2B=3 => A=3-2B
-4A-3B+2C=-25
4A-2B+C=43
From 1st eq insert into the 2nd and 3rd:
-4(3-2B)-3B+2C=-25
4(3-2B)-2B+C=43
-12+8B-3B+2C=-25
12-8B-2B+C=43
5B+2C=-13
-10B+C=31
5B+2C=-13
-20B+2C=62
25B=-75 => B=-3
C=31+10B=31-30=1 => C=1
A=3-2B=3+6=9 => A=9
So, the integral is broken apart:
int9/(2x+1)dx+int(-3)/(x-2)dx+int1/(x-2)^2dx=I
2x+1=t, 2dx=dt, dx=dt/2,
x-2=u, dx=du,
x-2=m, dx=dm
I=int9/tdt/2+int(-3)/udu+int1/m^2dm=
=9/2intdt/t-3int(du)/u+intm^-2dm=
=9/2lnabs(t)-3lnabs(u)+m^-1/-1+C=
=9/2lnabs(2x+1)-3lnabs(x-2)-1/(x-2)+C