#intx^2003/(1+x^2)^2002dx#
Let #u=1+x^2# so #du=2xdx#. Noting that this implies #x^2=u-1#:
#=1/2int((x^2)^1001(2x))/(1+x^2)^2002dx=1/2int(u-1)^1001/u^2002du#
Rewrite the numerator using the binomial theorem, which states that #(x+y)^n=sum_(k=0)^n((n),(k))x^(n-k)y^k#:
#=1/2int1/u^2002(sum_(k=0)^1001((1001),(k))u^(1001-k)(-1)^k)du#
Every #u# term in the series will have its exponent reduced by #2002#:
#=1/2int(sum_(k=0)^1001((1001),(k))(-1)^ku^(-1001-k))du#
Which can be written as:
#=1/2sum_(k=0)^1001(-1)^k((1001),(k))intu^(-1001-k)du#
Using the rule #intt^ndt=t^(n+1)/(n+1)#:
#=1/2sum_(k=0)^1001(-1)^k((1001),(k))u^(-1000-k)/(-1000-k)#
Rearranging slightly and adding a constant of integration:
#=C+1/2sum_(k=0)^1001((1001),(k))(-1)^(k+1)/(u^(k+1000)(k+1000))#
#=C+1/2sum_(k=0)^1001((1001),(k))(-1)^(k+1)/((1+x^2)^(k+1000)(k+1000))#
