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

2 Answers
Jun 11, 2017

#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#

Jun 12, 2017

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"#