`int (2x^3-4x^2-15x+5)/(x^2-2x-8) dx` Use partial fractions to find the indefinite integral

`int(2x^3-4x^2-15x+5)/(x^2-2x-8)dx`

The integrand is a improper rational function,as the degree of the numerator is greater than the degree of the denominator.So we have to carry out division.

`(2x^3-4x^2-15x+5)/(x^2-2x-8)=2x+(x+5)/(x^2-2x-8)`

Since the polynomials do not completely divide, we have to continue with the partial fractions on the remainder and factor out the denominator.

`(x+5)/(x^2-2x-8)=(x+5)/(x^2-4x+2x-8)`

`=(x+5)/(x(x-4)+2(x-4))`

`=(x+5)/((x-4)(x+2))`

Now we create the partial fraction template,

`(x+5)/((x-4)(x+2))=A/(x-4)+B/(x+2)`

Multiply the above equation with the denominator,

`=>(x+5)=A(x+2)+B(x-4)`  

`=>x+5=Ax+2A+Bx-4B`

`=>x+5=(A+B)x+2A-4B`

Equating the...

`int(2x^3-4x^2-15x+5)/(x^2-2x-8)dx`

The integrand is a improper rational function,as the degree of the numerator is greater than the degree of the denominator.So we have to carry out division.

`(2x^3-4x^2-15x+5)/(x^2-2x-8)=2x+(x+5)/(x^2-2x-8)`

Since the polynomials do not completely divide, we have to continue with the partial fractions on the remainder and factor out the denominator.

`(x+5)/(x^2-2x-8)=(x+5)/(x^2-4x+2x-8)`

`=(x+5)/(x(x-4)+2(x-4))`

`=(x+5)/((x-4)(x+2))`

Now we create the partial fraction template,

`(x+5)/((x-4)(x+2))=A/(x-4)+B/(x+2)`

Multiply the above equation with the denominator,

`=>(x+5)=A(x+2)+B(x-4)`  

`=>x+5=Ax+2A+Bx-4B`

`=>x+5=(A+B)x+2A-4B`

Equating the coefficients of the like terms,

`A+B=1`   ---------------(1)

`2A-4B=5`  -------------(2)

Now we to solve the above linear equations to get the values of A and B,

Multiply equation 1 by 4,

`4A+4B=4`  -------------(3)

Now add equation 2 and 3,

`2A+4A=5+4`

`=>6A=9`

`=>A=9/6`

`=>A=3/2`

Plug in the value of A in equation 1 ,

`3/2+B=1`

`=>B=1-3/2`

`=>B=-1/2`

Plug in the values of A and B in the partial fraction template,

`(x+5)/((x-4)(x+2))=(3/2)/(x-4)+(-1/2)/(x+2)`

`=3/(2(x-4))-1/(2(x+2))`

`int(2x^3-4x^2-15x+5)/(x^2-2x-8)dx=int(2x+3/(2(x-4))-1/(2(x+2)))dx`

Apply the sum rule,

`=int2xdx+int3/(2(x-4))dx-int1/(2(x+2))dx`

Take the constant out,

`=2intxdx+3/2int1/(x-4)dx-1/2int1/(x+2)dx`

Apply the power rule for the first integral and use the common integral `int1/xdx=ln|x|` for the second and third integral:    

 `=2x^2/2+3/2ln|x-4|-1/2ln|x+2|`  

Simplify and a constant C to the solution,

`=x^2+3/2ln|x-4|-1/2ln|x+2|+C`