Evaluate the integral? : #int x^2/(x^2+1)^2 dx#

Let us denote the given integral by #I#:

# I = int \ x^2/(x^2+1)^2 dx #

We can simply the fractional integrand with a slight manipulation:

# I = int \ (x^2+1-1)/(x^2+1)^2 \ dx #
# \ \ = int \ (x^2+1)/(x^2+1)^2 - (1)/(x^2+1)^2 \ dx #
# \ \ = int \ 1/(x^2+1) \ dx - int \ (1)/(x^2+1)^2 \ dx #
# \ \ = arctanx - int \ (1)/(x^2+1)^2 \ dx + C#

For this next integral we can use a substitution: Let

# tan theta = x => x^2 +1 = tan^2 theta+1 = sec^2 theta #
# dx/(d theta)=sec^2 theta #

Substituting we see that:

# int \ (1)/(x^2+1)^2 \ dx = int \ (1)/(sec^2 theta)^2\ sec^2 theta \ d theta #
# " " = int \ (1)/(sec^2 theta)^2\ sec^2 theta \ d theta #
# " " = int \ (1)/(sec^2 theta) \ d theta #
# " " = int \ cos^2 theta \ d theta #
# " " = int \ 1/2(1+cos2theta) \ d theta #
# " " = 1/2 \ int \ 1+cos2theta \ d theta #
# " " = 1/2 (theta + 1/2sin2theta) #
# " " = 1/2 (theta + sinthetacostheta) #
# " " = 1/2 (theta + sinthetacostheta * costheta/costheta)#
# " " = 1/2 (theta + sintheta/costheta cos^2theta)#
# " " = 1/2 (theta + tan theta/sec^2theta)#
# " " = 1/2 (arctanx + x/(x^2+1) )#

Combining this with the earlier result gives:

# I = arctanx - 1/2 (arctanx + x/(x^2+1) ) + C #
# \ \ = 1/2arctanx - x/(2(x^2+1)) + C #