Question

How do you integrate `((3x^2)-25x+43)dx/((2x+1)((x-2)^2))`?

Answer 1

`9/2lnabs(2x+1)-3lnabs(x-2)-1/(x-2)+C`

Explanation:

Lets rewrite integrand `(3x^2-25x+43)/((2x+1)(x-2)^2)` as follows:

`(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`