Question

How do you integrate `int (3x^2-2x+5)/(x^2+1)^2` using partial fractions?

Answer 1

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

Explanation:

Write the numerator as:

`3x^2-2x+5 = 3(x^2+1) -2x +2`

so that:

`int (3x^2-2x+5)/(x^2+1)^2 dx = 3int dx/(1+x^2) - int (2xdx)/(x^2+1)^2 +2 int dx/(x^2+1)^2`

Now solve the integral separately:

`int dx/(1+x^2) = arctanx + C_1`

`int (2xdx)/(x^2+1)^2 = int (d(x^2+1))/(x^2+1)^2 = -1/(x^2+1)+C_2`

The third integral can be resolved by substituting:

`x=tant`

`dx = dt/cos^2t`

`int dx/(x^2+1)^2 = int dt/(cos^2t(1+tan^2t)^2)`

use now the identity:

`1+tan^2t = 1/cos^2t`

`int dx/(x^2+1)^2 = int dt/(cos^2t/cos^4t) = int cos^2tdt`

and as `cos^2t = (1+cos2t)/2`

`int dx/(x^2+1)^2 = 1/2 int dt + 1/2 int cos(2t)dt`

`int dx/(x^2+1)^2 = t/2 + 1/4 int cos(2t)d(2t)`

`int dx/(x^2+1)^2 = t/2 + sin(2t)/4+C_3`

Use now the parametric formula:

`sin(2t) = (2tant)/(1+tan^2t)`

to undo the substitution:

`int dx/(x^2+1)^2 = arctanx/2 + 1/2x/(1+x^2)+C_3`

Putting it together:

`int (3x^2-2x+5)/(x^2+1)^2 dx = 3arctanx +1/(x^2+1) + 2(arctanx/2 + 1/2x/(1+x^2))+C`

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

Answer 2

This answer will only look at the partial fraction decomposition of the integrand as its integral has already been evaluated in the other answer.

Explanation:

`(3x^2-2x+5)/(x^2+1)^2-=(A+Bx)/(x^2+1)+(C+Dx)/(x^2+1)^2`

`3x^2-2x+5=(A+Bx)(x^2+1)+C+Dx`

`3x^2-2x+5=A+Ax^2+Bx^3+(B+D)x+C`

Compare coefficients of all `x`-terms

`5=A+C` (1)

`-2=B+D` (2)

`3=A` (3)

`0=B` (4)

Sub (4) into (2) and (3) into (1)

`-2=D`

`5=3+C rArr C=2`

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

This can be rewritten as:

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

`int3/(x^2+1)-(2(x-1))/(x^2+1)^2` `dx= 4arctanx + (x+1)/(x^2+1) +"c"`