`int (x^2 - 5x + 16)/((2x + 1)(x - 2)^2) dx` Evaluate the integral

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gsarora17 eNotes educator| Certified Educator

`int(x^2-5x+16)/((2x+1)(x-2)^2)dx2`

Let's first express the integrand as sum of proper rational expressions by applying partial fraction decomposition,

`(x^2-5x+16)/((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^2-4x+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)`

`=(x^2(A+2B)+x(-4A-3B+2C)+4A-2B+C)/((2x+1)(x-2)^2)`

Now equate the coefficients of the polynomial in the numerator on the both sides,

`A+2B=1` --------------------------------(1)

`-4A-3B+2C=-5`  -----------------(2)

`4A-2B+C=16`  ----------------------(3)

Now let's solve the above three equations by the method of substitution,

From equation 1 :`A=1-2B`

Substitute the above value of A in equation 2 ,

`-4(1-2B)-3B+2C=-5`

`-4+8B-3B+2C=-5`

`5B+2C=-5+4`

`5B+2C=-1`     ------------------------ (4)

Now substitute the value of A in equation 3,

`4(1-2B)-2B+C=16`

`4-8B-2B+C=16`

`4-10B+C=16`

`-10B+C=16-4`

`-10B+C=12`    -----------------------(5)

Now solve the equations 4 and 5 by the method of elimination,

Multiply equation 4 by 2,

`10B+4C=-2`  ----------------------(6)

Now add the equations 5 and 6,

`5C=12-2=10`

`C=10/5=2`

Plug the value of C in equation 5.

`-10B+2=12`

`-10B=12-2=10`

`B=10/-10=-1`

Plug the value of B in equation 1.

`A+2(-1)=1`

`A-2=1`

`A=1+2=3`

`:.int(x^2-5x+6)/((2x+1)(x-2)^2)dx=int(3/(2x+1)+(-1)/(x-2)+2/(x-2)^2)dx`

`=3int1/(2x+1)dx-int1/(x-2)dx+2int1/(x-2)^2dx`

Now let's evaluate the above three integrals,

`int1/(2x+1)dx`

Let's apply the integral substitution:`u=2x+1`

`du=2dx`

`=int1/(2u)du`

`=1/2ln|u|`

Substitute back u=2x+1,

`=1/2ln|2x+1|`

Now let's evaluate `int1/(x-2)dx`

apply integral substitution: `v=x-2`

`dv=dx`

`=int1/vdv`

`=ln|v|`

substitute back v=x-2,

`=ln|x-2|`

Now let's evaluate integral `int1/(x-2)^2dx`

apply the integral substitution: t=x-2

`dt=dx`

`int1/t^2dt`

`=intt^-2dt`

`=t^(-2+1)/(-2+1)`

`=-1/t`

Substitute back t=x-2,

`=-1/(x-2)`

`:.int(x^2-5x+16)/((2x+1)(x-2)^2)dx=3(1/2ln|2x+1|)-1ln|x-2|+2(-1/(x-2))`

`=3/2ln|2x+1|-ln|x-2|-2/(x-2)+C`   where C is a constant

 

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