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`int (8x) / (x^3+x^2-x-1) dx` Use partial fractions to find the indefinite integral

`int(8x)/(x^3+x^2-x-1)dx`


`(8x)/(x^3+x^2-x-1)=(8x)/((x^3+x^2)-1(x+1))`


`=(8x)/((x^2(x+1)-1(x+1)))`


`=(8x)/((x+1)(x^2-1))`


`=(8x)/((x+1)(x+1)(x-1))`


`=(8x)/((x-1)(x+1)^2)`


Now let's form the partial fraction template,


`(8x)/((x-1)(x+1)^2)=A/(x-1)+B/(x+1)+C/(x+1)^2`


Multiply the equation by the denominator,


`8x=A(x+1)^2+B(x-1)(x+1)+C(x-1)`


`8x=A(x^2+2x+1)+B(x^2-1)+C(x-1)`


`8x=Ax^2+2Ax+A+Bx^2-B+Cx-C`


`8x=(A+B)x^2+(2A+C)x+A-B-C`


Comparing the coefficients of the like terms,


`A+B=0`      -----------------(1)


`2A+C=8`   -----------------(2)


`A-B-C=0`  ---------------(3)


From equation 1,


`B=-A`


Substitute B in equation 3,


`A-(-A)-C=0`


`2A-C=0`      ---------------(4)


Now add equations 2 and 4,


`4A=8`


`A=8/4`


`A=2`


`B=-A=-2`


Plug in the value of A in equation 4,


`2(2)-C=0`


`C=4`


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


`(8x)/((x-1)(x+1)^2)=2/(x-1)+(-2)/(x+1)+4/(x+1)^2`  


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


Apply the sum rule,


`=int2/(x-1)dx-int2/(x+1)dx+int4/(x+1)^2dx`


Take the constant out,


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


Now let's evaluate the above three integrals separately,


`int1/(x-1)dx`


Apply integral substitution:`u=x-1`


`du=1dx`


`=int1/udu`


use the common integral `int1/xdx=ln|x|`


`=ln|u|`


Substitute back `u=x-1`


`=ln|x-1|`


Now let's evaluate second integral,


`int1/(x+1)dx`


Apply integral substitution: `u=x+1`


`du=dx`


`=int1/udu`


`=ln|u|`


Substitute back `u=x+1`


`=ln|x+1|`


Now evaluate the third integral,


`int1/(x+1)^2dx`


apply integral substitution: `u=(x+1)`


`du=dx`


`=int1/u^2du`


`=intu^(-2)du`


apply power rule,


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


`=-1/u`


Substitute back `u=x+1`


`=-1/(x+1)`


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


Simplify and add a constant C to the solution,


`=2ln|x-1|-2ln|x+1|-4/(x+1)+c`

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