Question

How do you integrate `f(x)=(x^2-2x)/((x^2-3)(x-2)(x+8))` using partial fractions?

Answer 1

`=1/61{4ln|x^2-3|-8ln|x+8|-sqrt3/2ln|(x-sqrt3)/(x+sqrt3)|}+C`

Explanation:

Here,

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

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

`I=intx/((x^2-3)(x+8))dx`

Partial Fractions :

`x/((x^2-3)(x+8))=A/(x+8)+(Bx+C)/(x^2-3)`

`=>x=A(x^2-3)+(Bx+C)(x+8)`

`=>x=Ax^2-3A+Bx^2+8Bx+Cx+8C`

`=>x=x^2(A+B)+x(8B+C)+8C-3A`

Comparing coefficient of `x^2, x and`constant term :

`A+B=0to(1),8B+C=1 to(2), 8C-3A=0 to(3)`

From`(1)` , `B=-Ato(4) and` from `(3)` , `C=(3A)/8to(5)`

So, from `(2)` , `-8A+(3A)/8=1=>-(61A)=8=>A=-8/61`

From `(4)` , `B=8/61`

From `(5)` ,`C=3/8(-8/61)=>C=-3/61`

Hence,

`I=int[(-8/61)/(x+8)+((8/61)x-(3/61))/(x^2-3)]dx`

`=-8/61int1/(x+8)dx+8/61intx/(x^2-3)dx-3/61int1/(x^2-3)dx`

#=-8/61ln|x+8|+4/61color(red)(int(2x)/(x^2-3)dx)-3/61color(blue)(int1/(x^2- (sqrt3)^2)dx#

=#-8/61ln|x+8|+4/61color(red)(ln|x^2-3|)-3/61*color(blue)(1/(2sqrt3)ln|(x- sqrt3)/(x+sqrt3)|+C#

`=1/61{4ln|x^2-3|-8ln|x+8|-sqrt3/2ln|(x-sqrt3)/(x+sqrt3)|}+C`

Note:

`color(red)((1)int(f'(x))/(f(x))dx=ln|f(x)|+c`

`color(blue)((2)int1/(X^2-A^2)dX=1/(2A)ln|(X-A)/(X+A)|+c`

Answer 2

The answer is `=(8+sqrt3)/122ln(|x+sqrt3|)+(8-sqrt3)/122ln(|x-sqrt3|)-8/61ln(|x+8|)+C`

Explanation:

The function is

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

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

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

Perform the decomposition into partial fractions

`x/((x^2-3)(x+8))=(Ax+B)/(x^2-3)+C/(x+8)`

`=((Ax+B)(x+8)+(C)(x^2-3))/((x^2-3)(x+8))`

The denominators are the same, compare the numerators

`x=(Ax+B)(x+8)+(C)(x^2-3)`

Compare the LHS and the RHS

Coefficients of `x^2`, `=>`, `0=A+C`, `=>`, `A=-C`

Let `x=-8`, `=>`, `-8=61C`, `=>`, `C=-8/61`

`A=8/61`

`0=8B-3C`, `=>`, `B=3/8*-8/61=-3/61`

Therefore,

`x/((x^2-3)(x+8))=(8/61x-3/61)/(x^2-3)+(-8/61)/(x+8)`

The integral is

`int(xdx)/((x^2-3)(x+8))=1/61int((8x-3)dx)/(x^2-3)-8/61int(dx)/(x+8)`

`I=I_1-I_2`

`I_2=8/61int(8dx)/(x+8)=8/61ln(x+8)`

For the calculation of `I_2`, perform the decomposition into partial fractions.

`(8x-3)/(x^2-3)=A/(x+sqrt3)+B/(x-sqrt3)`

`=(A(x-sqrt3)+B(x+sqrt3))/(x^2-3)`

Comparing the numerators,

`8x-3=A(x-sqrt3)+B(x+sqrt3)`

Let `x=-sqrt3`, `=>`, `-8sqrt3-3=A*-2sqrt3`, `=>`, `A=(8sqrt3+3)/(2sqrt3)=4+1/2sqrt3`

Let `x=sqrt3`, `=>`, `8sqrt3-3=B*2sqrt3`, `=>`, `B=(8sqrt3-3)/(2sqrt3)=4-1/2sqrt3`

Therefore,

`(8x-3)/(x^2-3)=(4+1/2sqrt3)/(x+sqrt3)+(4-1/2sqrt3)/(x-sqrt3)`

So,

`I_1=1/61int((8x-3)dx)/(x^2-3)=1/61int((4+1/2sqrt3)dx)/(x+sqrt3)+1/61int((4-1/2sqrt3)dx)/(x-sqrt3)`

`=(4+1/2sqrt3)/61ln(x+sqrt3)+(4-1/2sqrt3)/61ln(x-sqrt3)`

`=(8+sqrt3)/122ln(x+sqrt3)+(8-sqrt3)/122ln(x-sqrt3)`

And finally,

`int(xdx)/((x^2-3)(x+8))=(8+sqrt3)/122ln(|x+sqrt3|)+(8-sqrt3)/122ln(|x-sqrt3|)-8/61ln(|x+8|)+C`