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

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Integrate `int(x^2-2x-1)/[(x-1)^2(x^2+1)]dx`

Rewrite the rational function using partial fractions.

`(x^2-2x-1)/[(x-1)^2(x^2+1)]=A/(x-1)+B/(x-1)^2+(Cx+D)/(x^2+1)` 

`x^2-2x-1=A(x-1)(x^2+1)+B(x^2+1)+(Cx+D)(x-1)^2`

`x^2-2x-1=A(x^3-x^2+x-1)+Bx^2+B+(Cx+D)(x^2-2x+1)`

 

`x^2-2x-1=Ax^3-Ax^2-Ax-A+Bx^2+B `

`+ Cx^3-2Cx^2+Cx+Dx^2-2Dx+D `

  

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

`+(A+C-2D)x+(-A+B+D)`

 

Equate coefficients and solve for A, B, C, and D.

`0=A+C`

`A=-C` 

`-2=A+C-2D`

`-2=-C+C-2D`

`-2=-2D`

`D=1`

 

`1=-A+B-2C+D`

`1=C+B-2C+1`

`0=-1C+B`

`B=C `

 

`-1=-A+B+D`

`-1=C+B+1`

`-2=B+B`

`-2=2B`

`B=-1`

 

`C=-1`

 

`A=1`

 

`int(x^2-2x-1)/[(x-1)^2(x^2+1)]dx`

`=int[1/(x-1)]dx-int[1/(x-1)^2]dx+int[(-1x+1)/(x^2+1)]dx`

`=int[1/(x-1)]dx-int[1/(x-1)^2]dx+int[-x/(x^2+1)]dx+int[1/(x^2+1)]dx`

 

The first integral follows the pattern `int(du)/u=ln|u|+C`

` `

`int[1/(x-1)]dx=ln|x-1|+C`

 

Integrate the second integral using u-substitution.

Let `u=x-1`

`(du)/(dx)=1`

`du=dx`

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

=`-intu^-2du`

`=1/u+C`

`1/(x-1)+C`

 

Integrate the third integral using u-subsitution.

Let `u=x^2+1`

` `

`(du)/(dx)=2x`

`(dx)=(du)/(2x)`

`-intx/(x^2+1)dx`

`=-int(x)/(u)*(du)/(2x)`

`=-1/2ln|u|+C`

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

 

The fourth integral matches the pattern

 `intdx/(x^2+a^2)=(1/a)tan^-1(x/a)+C`

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

`=tan^-1(x)+C`

 

The final answer is:

`ln|x-1|+1/(x-1)-1/2ln|x^2+1|+tan^-1(x)+C`

 

 

 

 

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